physical chemistry - Positive or Negative Anode/Cathode in Electrolytic/Galvanic Cell - Chemistry Stack Exchange Positive or Negative Anode/Cathode in Electrolytic/Galvanic Cell up vote 19 down vote favorite 5 In a galvanic (volatic) cell, the anode is considered negative and the cathode is considered positive. This seems reasonable as the anode is the source of electrons and and the cathode is where the electrons flow to. However, in an electrolytic cell, the anode is taken to be positive while the cathode is now negative. However, the reaction is still similar, whereby electrons from the anode flow to the positive terminal of the battery, and electrons from the battery flow to the cathode. So why does the sign of the cathode and anode switch when considering a electrolytic cell? The anode is the electrode where the oxidation reaction \begin{align} \ce{Red -> Ox + e-} \end{align} takes place while the cathode is the electrode where the reduction reaction \begin{align} \ce{Ox + e- -> Red} \end{align} takes place. That's how cathode and anode are defined.

Galvanic cell Now, in galvanic cell the reaction proceeds without an external potential helping it along. Since at the anode you have the oxidation reaction which produces electrons you get a build-up of negative charge in the course of the reaction until electrochemical equilibrium is reached. Thus the anode is negative. At the cathode on the other hand you have the reduction reaction which consumes electrons (leaving behind positive (metal) ions at the electrode) and thus leads to a build-up of positive charge in the course of the reaction until electrochemical equilibrium is reached. Thus the cathode is positive. Electrolytic cell In an electrolytic cell you apply an external potential to enforce the reaction to go in the opposite direction. Now the reasoning is reversed. At the negative electrode where you have produced a high electron potential via an external voltage source electrons are "pushed out" of the electrode, thereby reducing the oxidized species $\ce{Ox}$, because the electron energy level inside the electrode (Fermi Level) is higher than the energy level of the LUMO of $\ce{Ox}$ and the electrons can lower their energy by occupying this orbital - you have very reactive electrons so to speak.

So the negative electrode will be the one where the reduction reaction will take place and thus it's the cathode. At the positive electrode where you have produced a low electron potential via an external voltage source electrons are "sucked into" the electrode leaving behind the the reduced species $\ce{Red}$ because the electron energy level inside the electrode (Fermi Level) is lower than the energy level of the HOMO of $\ce{Red}$. So the positive electrode will be the one where the oxidation reaction will take place and thus it's the anode. A tale of electrons and waterfalls Since there is some confusion concerning the principles on which an electrolysis works, I'll try a metaphor to explain it. Electrons flow from a region of high potential to a region of low potential much like water falls down a water fall or flows down an inclined plane. The reason is the same: water and electrons can lower their energy this way. Now the external voltage source acts like two big rivers connected to waterfalls: one at a high altitude that leads towards a waterfall - that would be the minus pole - and one at a low altitude that leads away from a waterfall - that would be the plus pole.

The electrodes would be like the points of the river shortly before or after the waterfalls in this picture: the cathode is like the edge of a waterfall where the water drops down and the anode is like the point where the water drops into. Ok, what happens at the electrolysis reaction? At the cathode you have the high altitude situation. So the electrons flow to the "edge of their waterfall". They want to "fall down" because behind them the river is pushing towards the edge exerting some kind of "pressure". But where can they fall down to? The other electrode is seperated from them by the solution and usually a diaphragma. But there are $\ce{Ox}$ molecules that have empty states that lie energetically below that of the electrode. Those empty states are like small ponds lying at a lower altitude where a little bit of the water from the river can fall into. So every time such an $\ce{Ox}$ molecule comes near the electrode an electron takes the opportunity to jump to it and reduce it to $\ce{Red}$. But that does not mean that the electrode is suddenly missing an electron because the river is replacing the "pushed out" electron immediately.

And the voltage source (the source of the river) can't run dry of electrons because it gets its electrons from the power socket. Now the anode: At the anode you have the low altitude situation. So here the river lies lower than everything else. Now you can imagine the HOMO-states of the $\ce{Red}$ molecules as small barrier lakes lying at a higher altitude than our river. When a $\ce{Red}$ molecule comes close to the electrode it is like someone opening the floodgates of the barrier lake's dam. The electrons flow from the HOMO into the electrode thus creating an $\ce{Ox}$ molecule. But the electrons don't stay in the electrode, so to speak, they are carried away by the river. And since the river is such a vast entity (lots of water) and usually flows into an ocean, the little "water" that is added to it doesn't change the river much. It stays the same, unaltered, so that everytime a floodgate gets opened the water from the barrier lake will drop the same distance.

up vote 3 down vote I'm no expert nor scholar, but from what I am reading in all of these explanations, and what I notice from the illustration, it becomes obvious...at least to me...which I feel may clarify the polarity change between the Galvanic cell and electrolytic cell for this user. As established and understood, the source of electrons and transfer of ions flows from the negative pole, (Anode) and is received by the positive pole (Cathode) (intentionally using most basic terms) the anode is negative here because the the flow originates FROM the electrolyte, into the light bulb, for which, if the terminals of the bulb were labeled, they would match the electrolyte in the other cell as it is the force coming into the bulb pushing the flow to the cell's cathode, and the cell's cathode is pulling from the bulb. In the electrolytic cell, the "electrolyte" is taking the role of the light bulb of the Galvanic cell, since the electrons are being SENT TO it from the power source, and is not in itself the SOURCE of flow, but is SUBJECT TO the force from the source of flow.

SO just as the Galvanic cell's anode sends to the light bulb, and the electrolyte is labeled like the load of the galvanic cell, and transferring its incoming negative force from the current source, and this pushes through the electrolyte like the flow FROM the light bulb. It may be easier if you note that the SOURCE of power is NOT the electrolyte and technically, the black terminal of the power supply is the TRUE anode (Sending), and the red side the TRUE Cathode, (Receiving) but when identifying the reactive substance submerged/surrounded in the electrolytic substance, the anode is giving up its ions, which then add to the Cathode which is receiving them. Therefor the tags in the electrolytic cell are not naming the "source of flow", but the reaction of the substances involved, due TO the force/flow imposed on them from the power source, but is not THE source of power, and therefore should not be labeled AS one...and there are only two options for labeling them, and since it cannot be changed at the power source it can only b changed at the point of contact with the electrolyte! At least this is what I have come to understand by reviewing the comments and illustrations.

I sincerely hope it helps clarify the rationale for the reversal of labels for this user and any others struggling with the concept for being due to the the source of current having to be labeled as - Anode and + Cathode... forcing the object the current plays upon to be the opposite despite their poles and due to direction of flow. Daniel Electrochemical Cells In an electrochemical cell, we physically separate the oxidation and reduction chemistry in different "compartments". The electrons from the oxidation are then run through an external circuit before being used in the reduction reaction. As this is moving negative charge from one location to another, we need to compensate for this by moving other charges to balance this displacement of charge. This is accomplished by using a salt bridge that allows the migration of spectator ions to balance the flow of electrons. Each half of the electrochemical cell has an electrode to which the wire for our external circuit is connected. The chemistry takes place at the surface of this electrode. The electrode on the oxidation side is called the anode. The electrode on the reduction side is called the cathode. As a short hand, we simply refer to the two sides of the cell as the anode side and the cathode side.

For any given set of conditions, the chemistry is spontaneous in only one direction. When the cell is set up such that this is the direction we desire, we call the cell a voltaic cell (or a galvanic cell or a battery). When we want the chemistry to go in the non-spontaneous direction, the cell is an electrolytic cell and requires an external power source to drive the chemistry in a non-spontaneous direction. It is important to note that this is one of the amazing things about electrochemistry. We can use an external voltage to change the direction of a chemical reaction by adjusting the free energy (we essentially get to chose which is lower in free energy the products or reactants). So with an electrochemical cell, we can get a chemical reaction to move in both the spontaneous and the non-spontaneous direction. You are all familiar with this concept but you might not have thought about it as chemistry. When you are using a battery you are running a voltaic electrochemical cell in the spontaneous direction. When it gets to equilibrium the voltage on the battery is zero and no more current flows.

However, you can "recharge" the battery by attaching it to an external power supply and forcing the chemistry to go in the reverse direction until you have regenerated the original chemical contents. For more practice with electrochemical cell problems, see the worksheet and key below. Standard Reduction Potentials at 25 C TABLE Electrochemical Cell Parts In an electrochemical cell, we are running a redox reaction, but we have physically separated the oxidation and reduction reactions into different locations. These take place in different "halves" of the cell. We then connect the two halves of the cell with an external wire to allow for electron flow from one side to the other. We also connect the two halves with a "salt bridge" to allow spectator ions to flow to counter balance the electron flow. The oxidation and reduction reactions are now in different locations. To allow the electrons to flow from the oxidation reaction (that is producing electrons) to the reduction reaction (that is consuming the electrons), we need to connect the two reactions electrically. This is accomplished by using metal electrodes. These electrodes serve a the place where the chemistry is taking place. Sometimes they are part of the reaction. Sometimes they are simply used as a means to "deliver/collect" the electrons.

The electrode in the oxidation reaction is call the anode. The electrode for the reduction reaction is called the cathode. The electrons flow from the anode to the cathode. Let's look at a picture of the cell below which contains zinc metal and zinc ions on the oxidation side and copper metal and copper ions on the reduction side. The left hand side is a picture of the anode half of the cell where the oxidation is taking place. The solid piece of zinc is the anode. The righthand side is the reduction half where the copper 2+ ions are being reduced to copper metal. The solid piece of copper is the cathode. The two sides are connected by a wire that allows electrons to flow form the anode to the cathode. They are also connected by a salt bridge. Since negative charge is flowing from the anode side to the cathode side, some spectator ions need to compensate this flow (or the cathode side will end up with lots of extra negative charge). So in the salt bridge either sodium + ions could flow from the anode side to the cathode side or sulfate 2- ions could flow from the cathode side to the anode side. Or both could happen.

For some reactions, neither the reactants nor the products are a metal. In this case, an inert (non-reactive) metal electrode is required to deliver/collect the electrons for the reaction. For example, in the cell below at the anode, Fe2+ is oxidized to Fe3+. This reaction occurs at the surface of a Pt electrode. Cell Short Hand It becomes very cumbersome to draw a picture of an electrochemical cell each time you would like to discuss one. As such, we have developed a short hand notation for a cell. For the cell above we have the anode on the left and the cathode on the right.

The oxidation is occurring at the anode \[\rm{Ni(s) \rightarrow Ni^{2+}(aq) + 2e^-}\] The reduction is occurring at the cathode \[\rm{2H^+(aq) + 2e^- \rightarrow H_2(g)}\] The overall reaction is \[\rm{Ni(s) + 2H^+(aq) \rightarrow Ni^{2+}(aq) + H_2(g)}\] Rather than having this picture of the cell, we have a shorthand that is the same moving from left to right in our diagram. We start with the anode which we write to the right of a "" symbol. In total, this will look like < Ni | Ni2+ || H+ | H2 | Pt > This is the most generic of notations. To be more specific we might include the concentrations (or pressures) along with the phases.

< Ni(s) | Ni2+ (aq,1M) || H+ (aq, 1M) | H2 (g, 1 atm) | Pt(s) > Reading this from left to right, we have a nickel anode that is oxidizing to Ni2+ connected by a salt bridge to a solution of H+ that is reducing to H2 gas at a Pt cathode. Note: on both the anode and cathode sides of the notation the reactants are on the left and the products are on the right. (Ni -> Ni2+) (H+ -> H2). Also, in the cell notation the half reaction might not be balanced with each other or themselves. For example you write H+ not 2H+ because if you have a concentration, it is for H+. Standard Hydrogen Electrode For any half reaction, we can measure the potential (free energy) by comparing it to a standard reaction. The reaction we choose is the following. \[\rm{2H^+(aq) + 2e^- \rightarrow H_2(g)}\] Where the concentration of the H+ is 1M and the pressure of the H2 gas is 1 atm.

Such an electrode is called a "standard hydrogen electrode" or SHE. We can then make a cell with any other half reaction (under standard conditions where concentrations are 1M) and measure the potential (voltage). For example here is a cell to compare the potential of the reaction of Ni to Ni2+ to the SHE. We can then measure all such combinations and tabulate the data. Note: for some reactions, we are measuring the oxidation and for others the reduction, but we will always tabulate the reduction potential for the half reaction. Standard Reduction Potentials Having compared many reactions to the standard hydrogen potential, we can now make a table of reduction potentials for all half-reaction, (or oxidation potentials but we need to pick one and stick to it). Below is a table for reduction potentials. At the top of the table are the reactions that are "easiest" to oxidize and thus the hardest to run as reductions. At the bottom are the easiest reductions, and thus the most difficult to run as oxidations. When looking at the table, we need to be careful since everything is written as a reduction. For example, from this table we can find the substance that is easiest to oxidize. That is from the top of the table.

But the substance that is being oxidized appears as a product in this table. Li(s) is the easiest to oxidize. The easiest to reduce is a reactant in the table. F2 gas is the easiest to reduce. Since Li is easy to oxidize, it is an excellent reducing agent (it reduces something else when it is oxidized). F2 is a great oxidizing agent (it oxidizes something else when it is reduced). From this table, we can now figure out what reactions will be spontaneous. For example, if something is higher in the table (lower potential) it will be oxidized while the reactions with higher (more positive potentials) will be reduced. Will Ag+ oxidize Fe? Yes. How do we know? The reduction potential for Ag+ is more positive than that for Fe2+. So Ag+ is a strong enough oxidizing agent to oxidize Fe to Fe2+. On the other hand it could not oxidize H2O to H+ and O2. Below is as more extensive table of standard reduction potentials. An even larger data set can be found on the wikipedia link below. A pdf document of the Standard Reduction Potential Table used in class can be found on the attachment below.

Attachments … ode_potential_(data_page) - Standard Cell Potentials We can calculate the standard potential for any electrochemical cell from the standard potentials of the two half reactions. For example, imagine we had the following cell. < Ni | Ni2+ || Fe3+ , Fe2+ | Pt > We have at the anode side the reaction \[\rm{Ni \rightarrow Ni^{2+} + 2e^-}\] and at the cathode side the reaction \[\rm{Fe^{3+} + 1 e^- \rightarrow Fe^{2+}}\] The standard potential for this cell is simply \[\rm{\mathcal{E}^{\circ} = \mathcal{E}^{\circ}_{cathode} - \mathcal{E}^{\circ}_{anode}}\] Where \(\mathcal{E}^{\circ}\) for each electrode is the standard reduction potential for the half-reaction.

Using our table of standard reduction potentials we find for \[\rm{Ni^{2+} + 2e^- \rightarrow Ni(s) \mathcal{E}^{\circ} = -0.25 V}\] \[\rm{Fe^{3+} + e^- \rightarrow Fe^{2+} \mathcal{E}^{\circ} = +0.77 V}\] So the standard potential is \[\rm{\mathcal{E}^{\circ} = \mathcal{E}^{\circ}_{cathode} - \mathcal{E}^{\circ}_{anode} = 0.77 - (-0.25) = +1.02 V }\] VERY IMPORTANT. Note: the potential is simply the energy difference between the two half reactions. Do not try to multiple the potentials by the number of electrons. The number of electrons simply relates how many electrons there are per reaction. How many Fe3+ will be reduced per Ni atom that is oxidized. The potential difference (free energy difference) between the two half-reactions is not dependent on the number of electrons.

Voltaic Cells An electrochemical cell in which the chemistry is spontaneous is called a voltaic cell. This means that the oxidation will occur spontaneously at the anode and the reduction spontaneously at the cathode. We should note that the notation of something as a voltaic cell is a choice. Just as we choose what we want to call the "reactants" and "products" in a chemical reaction, our choice of anode and cathode is based upon what chemistry we would like to see occur. For a voltaic cell our choice ends up being the spontaneous choice. The standard potential for a voltaic cell is positive. The standard free energy for a reaction ΔrG0 is related to the standard potential, E0, such that negative free energy (spontaneous) corresponds to positive potential. The beauty of electrochemistry and electrochemical cells is that we can now directly measure the free energy difference by measuring electrical potential. Voltaic cells can be referred to as Galvanic Cells. This is a different word for the identical concept. They are also batteries. Since a battery is an electrochemical cell that produces a voltage (and current) spontaneously, it is a voltaic cell. You will find these three terms used interchangeably.

Below is a picture that summarize the ideas for a voltaic cell. For a voltaic cell E > 0, ΔG < 0. Electrons flow from anode to cathode. The one odd detail is the "sign" of the electrodes. For a voltaic cell the cathode is assigned the "+" sign. Electrolytic Cells An electrochemical cell in which the chemistry is non-spontaneous is called a electrolytic cell. This means that the oxidation will not occur spontaneously at the anode and the reduction will not be spontaneous at the cathode. This means we would like chemistry to occur that is uphill in free energy. This means that it can't happen unless we make it happen. For most chemical reactions, this can only be accomplished by coupling it to another reaction in which the free energy change is sufficiently negative that the two reactions together are spontaneous. In the case of electrochemistry, it is significantly easier as we can simply drive the reaction by attaching an external power supply. The standard potential for a voltaic cell is negative. The standard free energy for a reaction ΔrG0 is related to the standard potential, E0, such that positive free energy (non-spontaneous) corresponds to negative potential.

To make the reaction "go," we must apply a voltage that is significant enough to overcome the negative potential of the cell itself. Electrolytic cells thankfully don't have many names. However processes such as electroplating, or electrodeposition are typically electrolytic cells. Below is a picture that summarizes the cell nomenclature for a electrolytic cell. For an electrolytic cell E < 0, ΔG > 0. Electrons flow from anode to cathode (this is always the case). For an electrolytic cell however, this flow is not spontaneous but must be driven by an external power source. In an electrolytic cell, the anode has the "+" sign. © 2013 mccord/vandenbout/labrake Electrolytic Cells Electrolytic Cells Faraday's Law Voltaic cells use a spontaneous chemical reaction to drive an electric current through an external circuit. These cells are important because they are the basis for the batteries that fuel modern society. But they aren't the only kind of electrochemical cell. It is also possible to construct a cell that does work on a chemical system by driving an electric current through the system. These cells are called electrolytic cells.

Electrolysis is used to drive an oxidation-reduction reaction in a direction in which it does not occur spontaneously. The Electrolysis of Molten NaCl An idealized cell for the electrolysis of sodium chloride is shown in the figure below. A source of direct current is connected to a pair of inert electrodes immersed in molten sodium chloride. Because the salt has been heated until it melts, the Na+ ions flow toward the negative electrode and the Cl- ions flow toward the positive electrode. When Na+ ions collide with the negative electrode, the battery carries a large enough potential to force these ions to pick up electrons to form sodium metal. Negative electrode (cathode): Na+ + e- Na Cl- ions that collide with the positive electrode are oxidized to Cl2 gas, which bubbles off at this electrode. Positive electrode (anode): 2 Cl- Cl2 + 2 e- The net effect of passing an electric current through the molten salt in this cell is to decompose sodium chloride into its elements, sodium metal and chlorine gas. Electrolysis of NaCl: 2 NaCl(l) 2 Na(l) + Cl2(g) This example also illustrates the difference between voltaic cells and electrolytic cells.

Voltaic cells use the energy given off in a spontaneous reaction to do electrical work. Electrolytic cells use electrical work as source of energy to drive the reaction in the opposite direction. The dotted vertical line in the center of the above figure represents a diaphragm that keeps the Cl2 gas produced at the anode from coming into contact with the sodium metal generated at the cathode. The function of this diaphragm can be understood by turning to a more realistic drawing of the commercial Downs cell used to electrolyze sodium chloride shown in the figure below. Chlorine gas that forms on the graphite anode inserted into the bottom of this cell bubbles through the molten sodium chloride into a funnel at the top of the cell. Sodium metal that forms at the cathode floats up through the molten sodium chloride into a sodium-collecting ring, from which it is periodically drained. The diaphragm that separates the two electrodes is a screen of iron gauze, which prevents the explosive reaction that would occur if the products of the electrolysis reaction came in contact. The feed-stock for the Downs cell is a 3:2 mixture by mass of CaCl2 and NaCl.

This mixture is used because it has a melting point of 580oC, whereas pure sodium chloride has to be heated to more than 800oC before it melts. The Electrolysis of Aqueous NaCl The figure below shows an idealized drawing of a cell in which an aqueous solution of sodium chloride is electrolyzed. Once again, the Na+ ions migrate toward the negative electrode and the Cl- ions migrate toward the positive electrode. But, now there are two substances that can be reduced at the cathode: Na+ ions and water molecules. Cathode (-): O2 + 4 H+ + 4 e- Eoox = -1.23 V The standard-state potentials for these half-reactions are so close to each other that we might expect to see a mixture of Cl2 and O2 gas collect at the anode. In practice, the only product of this reaction is Cl2. Anode (+): 2 Cl- Cl2 + 2 e- At first glance, it would seem easier to oxidize water (Eoox = -1.23 volts) than Cl- ions (Eoox = -1.36 volts).

It is worth noting, however, that the cell is never allowed to reach standard-state conditions. The solution is typically 25% NaCl by mass, which significantly decreases the potential required to oxidize the Cl- ion. The pH of the cell is also kept very high, which decreases the oxidation potential for water. The deciding factor is a phenomenon known as overvoltage, which is the extra voltage that must be applied to a reaction to get it to occur at the rate at which it would occur in an ideal system. Under ideal conditions, a potential of 1.23 volts is large enough to oxidize water to O2 gas. Under real conditions, however, it can take a much larger voltage to initiate this reaction. (The overvoltage for the oxidation of water can be as large as 1 volt.) By carefully choosing the electrode to maximize the overvoltage for the oxidation of water and then carefully controlling the potential at which the cell operates, we can ensure that only chlorine is produced in this reaction. In summary, electrolysis of aqueous solutions of sodium chloride doesn't give the same products as electrolysis of molten sodium chloride. Electrolysis of molten NaCl decomposes this compound into its elements.

2 NaCl(aq) + 2 H2O(l) 2 Na+(aq) + 2 OH-(aq) + H2(g) + Cl2(g) Because the demand for chlorine is much larger than the demand for sodium, electrolysis of aqueous sodium chloride is a more important process commercially. Electrolysis of an aqueous NaCl solution has two other advantages. It produces H2 gas at the cathode, which can be collected and sold. It also produces NaOH, which can be drained from the bottom of the electrolytic cell and sold. The dotted vertical line in the above figure represents a diaphragm that prevents the Cl2 produced at the anode in this cell from coming into contact with the NaOH that accumulates at the cathode. When this diaphragm is removed from the cell, the products of the electrolysis of aqueous sodium chloride react to form sodium hypo-chlorite, which is the first step in the preparation of hypochlorite bleaches, such as Chlorox.

Cl2(g) + 2 OH-(aq) 2 H2O(l) 2 H2(g) + O2(g) A pair of inert electrodes are sealed in opposite ends of a container designed to collect the H2 and O2 gas given off in this reaction. The electrodes are then connected to a battery or another source of electric current. By itself, water is a very poor conductor of electricity. We therefore add an electrolyte to water to provide ions that can flow through the solution, thereby completing the electric circuit. The electrolyte must be soluble in water. It should also be relatively inexpensive. Most importantly, it must contain ions that are harder to oxidize or reduce than water. 2 H2O + 2 e- Faraday's Law Faraday's law of electrolysis can be stated as follows. The amount of a substance consumed or produced at one of the electrodes in an electrolytic cell is directly proportional to the amount of electricity that passes through the cell. In order to use Faraday's law we need to recognize the relationship between current, time, and the amount of electric charge that flows through a circuit. By definition, one coulomb of charge is transferred when a 1-amp current flows for 1 second.

1 C = 1 amp-s Example: To illustrate how Faraday's law can be used, let's calculate the number of grams of sodium metal that will form at the cathode when a 10.0-amp current is passed through molten sodium chloride for a period of 4.00 hours. We start by calculating the amount of electric charge that flows through the cell. Before we can use this information, we need a bridge between this macroscopic quantity and the phenomenon that occurs on the atomic scale. This bridge is represented by Faraday's constant, which describes the number of coulombs of charge carried by a mole of electrons. Thus, the number of moles of electrons transferred when 144,000 coulombs of electric charge flow through the cell can be calculated as follows. According to the balanced equation for the reaction that occurs at the cathode of this cell, we get one mole of sodium for every mole of electrons. Cathode (-): Sample Questions - Chapter 21 (e) none of the above 12.

For a voltaic (or galvanic) cell using Ag,Ag+ (1.0 M) and Zn,Zn2+ (1.0 M) half-cells, which of the following statements is incorrect? (a) The zinc electrode is the anode. (b) Electrons will flow through the external circuit from the zinc electrode to the silver electrode. (c) Reduction occurs at the zinc electrode as the cell operates. (d) The mass of the zinc electrode will decrease as the cell operates. (e) The concentration of Ag+ will decrease as the cell operates. 13. Consider the standard voltaic (or galvanic) cell: Fe,Fe2+ versus Au,Au3+. Which answer identifies the cathode and gives the Eo for the cell? (a) Fe, -0.44 V In voltaic cells, such as those diagrammed in your text, the salt bridge _______ .

(a) is not necessary in order for the cell to work (b) acts as a mechanism to allow mechanical mixing of the solutions (c) allows charge balance to be maintained in the cell (d) is tightly plugged with firm agar gel through which ions cannot pass (e) drives free electrons from one half-cell to the other 21. Which of the following statements is(are) true for all voltaic (or galvanic) cells? (I) Reduction occurs at the cathode. (II) The anode gains mass during discharge (note: this means operation of the cell.) (III) The voltage is less than or equal to zero. (a) only III Electrolytic Cells - Chemistry LibreTexts Electrolytic Cells Contributors Voltaic cells are driven by a spontaneous chemical reaction that produces an electric current through an outside circuit. These cells are important because they are the basis for the batteries that fuel modern society. But they are not the only kind of electrochemical cell. The reverse reaction in each case is non-spontaneous and requires electrical energy to occur.

Introduction The general form of the reaction can be written as: Reactants \(\rightleftharpoons\) Products + Electrical Energy \(\longleftarrow\) Non spontaneous It is possible to construct a cell that does work on a chemical system by driving an electric current through the system. These cells are called electrolytic cells, and operate through electrolysis. Electrolysis is used to drive an oxidation-reduction reaction in a direction in which it does not occur spontaneously by driving an electric current through the system while doing work on the chemical system itself, and therefore is non-spontaneous. Electrolytic cells, like galvanic cells, are composed of two half-cells--one is a reduction half-cell, the other is an oxidation half-cell. The direction of electron flow in electrolytic cells, however, may be reversed from the direction of spontaneous electron flow in galvanic cells, but the definition of both cathode and anode remain the same, where reduction takes place at the cathode and oxidation occurs at the anode. Because the directions of both half-reactions have been reversed, the sign, but not the magnitude, of the cell potential has been reversed.

Differences between Galvanic and an Electrolytic cells Electrolytic cells are very similar to voltaic (galvanic) cells in the sense that both require a salt bridge, both have a cathode and anode side, and both have a consistent flow of electrons from the anode to the cathode. However, there are also striking differences between the two cells. The main differences are outlined below: Figure 1: Electrochemical Cells. A galvanic cell (left) transforms the energy released by a spontaneous redox reaction into electrical energy that can be used to perform work. The oxidative and reductive half-reactions usually occur in separate compartments that are connected by an external electrical circuit; in addition, a second connection that allows ions to flow between the compartments (shown here as a vertical dashed line to represent a porous barrier) is necessary to maintain electrical neutrality. The potential difference between the electrodes (voltage) causes electrons to flow from the reductant to the oxidant through the external circuit, generating an electric current.

In an electrolytic cell (right), an external source of electrical energy is used to generate a potential difference between the electrodes that forces electrons to flow, driving a nonspontaneous redox reaction; only a single compartment is employed in most applications. In both kinds of electrochemical cells, the anode is the electrode at which the oxidation half-reaction occurs, and the cathode is the electrode at which the reduction half-reaction occurs. Table 1: Properties of Galvanic and Electrochemical Cells Electrochemical cell (Galvanic Cell) A Galvanic cell converts chemical energy into electrical energy. An electrolytic cell converts electrical energy into chemical energy. Here, the redox reaction is spontaneous and is responsible for the production of electrical energy. The redox reaction is not spontaneous and electrical energy has to be supplied to initiate the reaction. The two half-cells are set up in different containers, being connected through the salt bridge or porous partition. Both the electrodes are placed in a same container in the solution of molten electrolyte. Here the anode is negative and cathode is the positive electrode. The reaction at the anode is oxidation and that at the cathode is reduction.

Here, the anode is positive and cathode is the negative electrode. The reaction at the anode is oxidation and that at the cathode is reduction. The electrons are supplied by the species getting oxidized. They move from anode to the cathode in the external circuit. The external battery supplies the electrons. They enter through the cathode and come out through the anode. Electrolytic Cells To explain what happens in an electrolytic cell let us examine the decomposition of molten sodium chloride into sodium metal and chlorine gas. The reaction is written below. Cl2 (g) <--------- Spontaneous ( electrochemical cell ) If molten \(NaCl_{(l)}\) is placed into the container and inert electrodes of \(C_{(s)}\) are inserted, attached to the positive and negative terminals of a battery, an electrolytic reaction will occur. Electrons from the negative terminal travel to the cathode and are used to reduce sodium ions into sodium atoms. The sodium will plate onto the cathode as it forms. The sodium ion are migrating towards the cathode.

\[ Na^+ + e^- \rightleftharpoons Na_{(s)}\] The negative Chlorine ions migrate towards the anode and release electrons as they oxidize to form chlorine atoms. The chlorine atoms will combine together to form chlorine gas which will bubble away. \[ 2Cl^- \rightleftharpoons Cl_{2 (g)} + 2e^-\] Note that the site of oxidation is still the anode and the site of reduction is still the cathode, but the charge on these two electrodes are reversed. Anode is now positive charged and the cathode has a negative charged. The conditions under which the electrolyte cell operates are very important. The substance that is the strongest reducing agent (the substance with the highest standard cell potential value in the table) will undergo oxidation. The substance that is the strongest oxidizing agent will be reduced. If an aqueous solution of sodium chloride were used in the above system, hydrogen would undergo reduction instead of sodium, because it is a stronger oxidizing agent that sodium.

Predicting Electrolysis Reaction There are four primary factors that determine whether or not electrolysis will take place even if the external voltage exceeds the calculated amount: An overpotential or voltage excess is sometimes needed to overcome interactions at the electrode surface. This case happens more frequently with gases. E.g. H2 (g) requies a 1.5 V overpotential, while Pt (s) requires 0 V overpotential There might be more than one electrode reaction that occurs meaning that there may be more than one half-reaction leaving two or more possibilities for the cell reaction. The reactants may be in nonstandard conditions which means that the voltage for the half cells may be less or more than the standard condition amount. For Example: Concentration of chloride ion = 5.5M not the unit activity of 1M. This means that the reduction of chloride = 1.31V not 1.36V The standard condition is to have a pH of 4 in the anode half cell but sometimes during nonstandard states, the pH may be higher or lower changing the voltage.

An inert electrode’s ability to electrolysis depend on the reactants in the electrolyte solution while an active electrode can run on its own to perform the oxidation or reduction half reaction. If all four of these factors are accounted for, we can successfully predict electrode half reactions and overall reactions in electrolysis. Exercise Predict the electrode reactions and the overall reaction when the anode is made of (a) copper and (b) platinum. Quantitative Aspects of Electrolysis Michael Faraday discovered in 1833 that there is always a simple relationship between the amount of substance produced or consumed at an electrode during electrolysis and the quantity of electrical charge Q which passes through the cell. For example, the half-equation \[Ag^+ + e^– \rightarrow Ag\] tells us that when 1 mol Ag+ is plated out as 1 mol Ag, 1 mol e– must be supplied from the cathode . Since the negative charge on a single electron is known to be 1.6022 × 10–19 C, we can multiply by the Avogadro constant to obtain the charge per mole of electrons.

This quantity is called the Faraday Constant, symbol F: F = 1.6022 × 10–19 C × 6.0221 × 1023 mol–1 = 9.649 × 104 C mol–1 Thus in the case of Eq. (1), 96 490 C would have to pass through the cathode in order to deposit 1 mol Ag. For any electrolysis the electrical charge Q passing through an electrode is related to the amount of electrons ne– by \[F=\dfrac{Q}{n_{e^-}}\] Thus F serves as a conversion factor between \(n_{e^-}\) and \(Q\). Often the electrical current rather than the quantity of electrical charge is measured in an electrolysis experiment.

Since a coulomb is defined as the quantity of charge which passes a fixed point in an electrical circuit when a current of one ampere flows for one second, the charge in coulombs can be calculated by multiplying the measured current (in amperes) by the time (in seconds) during which it flows: \[Q = It\] In this equation I represents current and t represents time. If you remember that coulomb = 1 ampere × 1 second 1 C = 1 A s you can adjust the time units to obtain the correct result. Now that we can predict the electrode half-reactions and overall reactions in electrolysis, it is also important to be able to calculate the quantities of reactants consumed and the products produced.

For these calculations we will be using the Faraday constant: 1 mol of electron = 96,485 C charge (C) = current (C/s) x time (s) (C/s) = 1 coulomb of charge per second = 1 ampere (A) Simple conversion for any type of problem: Convert any given time to seconds Take the current given (A) over the seconds, [1 c = (A)/(s)] Finally use the stoichiometry conversion of 1 mol of electron = 96,485 C (Faraday's Constant) Example 1 The electrolysis of dissolved Bromine sample can be used to determine the amount of Bromine content in sample. At the cathode, the reduction half reaction is \[Br^{2+}_{(aq)} + 2 e^- \rightarrow 2 Br^-\]. What mass of Bromine can be deposited in 3.00 hours by a current of 1.18 A?

SOLUTION: 3.00 hours x 60 min/hour x 60 sec/1 min x 1.18 C(A) / 1 sec x 1 mol e-/96,485 C = 0.132 mol e- 1) Predict the products of electrolysis by filling in the graph: Cl-, Br-, I-, H+, OH-, Cu2+, Pb2+, Ag+, K+, Na+, 2) Calculate the quantity of electrical charge needed to plate 1.386 mol Cr from an acidic solution of K2Cr2O7according to half-equation H2Cr2O7(aq) + 12H+(aq) + 12e– → 2Cr(s) + 7 H2O(l) 3) Hydrogen peroxide, H2O2, can be manufactured by electrolysis of cold concentrated sulfuric acid.

The reaction at the anode is 2H2SO4 → H2S2O8 + 2H+ + 2e– When the resultant peroxydisulfuric acid, H2S2O8, is boiled at reduced pressure, it decomposes: 2H2O + H2S2O8 → 2H2SO4 + H2O2 Calculate the mass of hydrogen peroxide produced if a current of 0.893 flows for 1 hour. 4) The electrolysis of dissolved Cholride sample can be used to determine the amount of Chloride content in sample. At the cathode, the reduction half reaction is Cl2+(aq) + 2 e- -> 2 Cl-. What mass of Chloride can be deposited in 6.25 hours by a current of 1.11 A? 5) In an electrolytic cell the electrode at which the electrons enter the solution is called the ______ ; the chemical change that occurs at this electrode is called _______. Electrochemical Cells Electrochemical Cells An extremely important class of oxidation and reduction reactions are used to provide useful electrical energy in batteries.

A simple electrochemical cell can be made from copper and zinc metals with solutions of their sulfates. In the process of the reaction, electrons can be transferred from the zinc to the copper through an electrically conducting path as a useful electric current. An electrochemical cell can be created by placing metallic electrodes into an electrolyte where a chemical reaction either uses or generates an electric current. Electrochemical cells which generate an electric current are called voltaic cells or galvanic cells, and common batteries consist of one or more such cells. In other electrochemical cells an externally supplied electric current is used to drive a chemical reaction which would not occur spontaneously. Such cells are called electrolytic cells . Go Back Voltaic Cells An electrochemical cell which causes external electric current flow can be created using any two different metals since metals differ in their tendency to lose electrons. Zinc more readily loses electrons than copper, so placing zinc and copper metal in solutions of their salts can cause electrons to flow through an external wire which leads from the zinc to the copper. Add annotation to illustration As a zinc atom provides the electrons, it becomes a positive ion and goes into aqueous solution, decreasing the mass of the zinc electrode.

On the copper side, the two electrons received allow it to convert a copper ion from solution into an uncharged copper atom which deposits on the copper electrode, increasing its mass. The two reactions are typically written Zn(s) -> Zn2+(aq) + 2e- Cu2+(aq) + 2e- -> Cu(s) The letters in parentheses are just reminders that the zinc goes from a solid (s) into a water solution (aq) and vice versa for the copper. It is typical in the language of electrochemistry to refer to these two processes as "half-reactions" which occur at the two electrodes. Zn(s) -> Zn2+(aq) + 2e- The zinc "half-reaction" is classified as oxidation since it loses electrons . The terminal at which oxidation occurs is called the "anode". For a battery, this is the negative terminal. The copper "half-reaction" is classified as reduction since it gains electrons . The terminal at which reduction occurs is called the "cathode". For a battery, this is the positive terminal.

Cu2+(aq) + 2e- -> Cu(s) In order for the voltaic cell to continue to produce an external electric current, there must be a movement of the sulfate ions in solution from the right to the left to balance the electron flow in the external circuit. The metal ions themselves must be prevented from moving between the electrodes, so some kind of porous membrane or other mechanism must provide for the selective movement of the negative ions in the electrolyte from the right to the left. Energy is required to force the electrons to move from the zinc to the copper electrode, and the amount of energy per unit charge available from the voltaic cell is called the electromotive force (emf) of the cell. Energy per unit charge is expressed in volts (1 volt = 1 joule/coulomb). Clearly, to get energy from the cell, you must get more energy released from the oxidation of the zinc than it takes to reduce the copper. The cell can yield a finite amount of energy from this process, the process being limited by the amount of material available either in the electrolyte or in the metal electrodes.

For example, if there were one mole of the sulfate ions SO42- on the copper side, then the process is limited to transferring two moles of electrons through the external circuit. The amount of electric charge contained in a mole of electrons is called the Faraday constant, and is equal to Avogadro's number times the electron charge: Faraday constant = F = NAe = 6.022 x 1023 x 1.602 x 10-19 = 96,485 Coulombs/mole The energy yield from a voltaic cell is given by the cell voltage times the number of moles of electrons transferred times the Faraday constant. Electrical energy output = nFEcell The cell emf Ecell may be predicted from the standard electrode potentials for the two metals. For the zinc/copper cell under the standard conditions, the calculated cell potential is 1.1 volts.

Question: What name is given to the negative electrode of an electrolytic cell?
Answer: Cathode