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Q.  What is the relationship between the standard EMF of the cell and the equilibrium constant of the cell reaction at 298  _K_?

Q.  Why is it not possible to measure the voltage of an isolated reduction half reaction or the potential of a single electrode?

Q.  State the factors that influence the value of cell potentialof the following cell: $M g(s)\left|M g^{2+}(a q) \| A g^{+}(a q)\right| A g(s)$

Q.  How is an electrochemical cell represented by cell notation?

Anode | Anode electrolyte || Cathode electrolyte | Cathode

Q.  After sometime, the voltage of an electrochemical cell becomes zero. Comment on the statement

Q.  What is over voltage?

Q.  Which allotropic form of carbon is used for making electrodes?

Q.  Write Nernst equation for the following cell reaction:$Z n(s)\left|Z n^{2+}(a q) \| \quad C u^{2+}(a q)\right| C u(s)$

Q.  Can a solution of _$\mathrm{ZnSO}_{4}$_ be stored in a vessel made up of copper?

Q.  What are the units of molar conductivity?

Q.  What is the relationship between specific conductance and equivalent conductance?

of the substance.

Q.  Why is not possible to determine $\wedge_{m}^{\infty}$ for weak electrolytes by extrapolation?

Q.  How is $\wedge_{m}^{\infty}$ of weak electrolytes calculated according to Kohlrausch’s law?

$\wedge_{m}^{\infty} C H_{3} C O O H=\wedge_{m}^{\infty} C H_{3} C O O N a+\wedge_{m}^{\infty} H C l-\wedge_{m}^{\infty} N a C l$

Q.  Which is evolved at cathode when an aqueous solution of _NaCl_ is electrolysed?

Q.  What is the function of platinised platinum in the standard hydrogen electrode?

Q.  Which is the basic difference in a primary and secondary cell?

Q.  Write an expression to relate molar conductivity of an electrolyte to its degree of dissociation.

Q.  What will happen to the value of emf of the cellZn $|$ Zn $^{2+}(0.1 M) \| C u^{2+}(0.1 M) | C u,$ if the concentration of the electrolyte in the anode compartment is increased

Q.  What would happen if the protective tin coating over on iron bucket is broken in some places

Q.  Which solution will allow greater conductance of electricity, $1 M N a C l$ at $293 \mathrm{Kor} 1 \mathrm{MNaCl}$ at $323 \mathrm{K}$

Q.  Give one point of distinction between emf and potential difference

Q.  Define the following terms: Specific and equivalent conductance

Q.  On passing one faraday of charge, on gram mole of the substance is deposited on the cathode. Comment

Q.  Rusting of iron is quicker in saline water than in ordinary water. Give reason

Saline medium has extra salts such as sodium chloride dissolved in water. This means that it has a greater concentration of electrolyte than ordinary medium. The ions present will favour the formation of more electrochemical cells and will thus promote rusting or corrosion.

Q.  What is standard hydrogen electrode? Give the reaction that occurs at this electrode when it acts as positive electrode in an electrochemical cell

Q.  What is fuel cell? Give the electrode reactions of hydrogen oxygen fuel cells

The reactions taking place in hydrogen – oxygen fuel cell:At cathode : $O_{2}+4 H^{+}+4 e^{-} \longrightarrow 2 H_{2} O$At anode : $2 H_{2} \longrightarrow 4 H^{+}+4 e^{-}$Net reaction $: 2 \mathrm{H}_{2}+\mathrm{O}_{2} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}$

Q.  With the help of a graph explain why it is not easy todetermine $\lambda_{m}^{\infty}$ for a weak electrolyte by extrapolating the concentration $-$ molar conductance curve as for strong electrolytes

Obviously $\wedge_{m}^{\infty}$ means molar conductance at infinite dilution. This can only be obtained by extrapolating the curve. The point at $y$ -axis where it meets of extrapolation is the value of It is apparent from the graph that in case of weak electrolytes the curve does not meet $y$ -axis on extrapolation, hence the value of can not be determined to the contrary in case of strong electrolytes the curve does meet the y-axis on extrapolation.

Q.  Write the cell reactions which occur in lead storage battery (i) When the battery is in use and (ii) when the battery is on charging

At anode $: P b(s)+S O_{4}^{2-} \longrightarrow P b S O_{4}(s)+2 e^{-}$At cathode:$P b O_{2}+4 H^{+}+S O_{4}^{2-}+2 e^{-} \longrightarrow P b S O_{4}(s)+2 H_{2} O(l)$Net reaction$P b+P b O_{2}+4 H^{+}+2 S O_{4}^{2-} \longrightarrow 2 P b S O_{4}(s)+2 H_{2} O(l)$(ii) During charging of the battery, the above reactions are reversed

Q.  Explain why electrolysis of aqueous solution of $N a C l$ gives $H_{2}$ at cathode and $C l_{2}$ anode. Write overall reaction

$N a C l(a q .) \longrightarrow N a^{+}(a q .)+C l^{-}(a q .)$$H_{2} O(l) \rightarrow H^{+}(a q .)+O H^{-}(a q .)$At cathode: Both $N a^{+}$ and $H^{+}$ ions are present near the cathode. But the discharge potential of is lower than thatof ions. So ions are discharged in preference to $N a^{+}$ ions$H^{+}+e^{-} \longrightarrow H ; H+H \longrightarrow H_{2}(g)$Thus, $\mathrm{H}_{2}$ gas is liberated at the cathode and $\mathrm{Na}^{+}$ ions remain in the solutionAt the anode: Both $C l^{-}$ and $O H^{-}$ ions are present nearthe anode. As the discharge potential of $C l^{-}$ ions is lowerthan that of $O H^{-}$ ions, so $C l^{-}$ ions are discharged inpreference to $\mathrm{OH}^{-}$ ions \\[ C l^{-}-e^{-} \longrightarrow C l^{\bullet} ; \quad C l^{\bullet}+C l^{\bullet} \longrightarrow C l_{2} \\]Thus $\mathrm{Cl}_{2}$ is liberated at anode and $\mathrm{OH}^{-}$ ions remain in the solution. The overall reaction is$N a C l(a q .)+H_{2} O(l) \longrightarrow N a^{+}(a q .)+O H^{-}(a q .)+1 / 2 H_{2}(g)$$+1 / 2 C l_{2}(g)$

Q.  What is corrosion? Describe the electrochemical phenomenon of rusting of Iron

water containing dissolved oxygen or $\mathrm{CO}_{2} .$ The pure iron acts as anode and impure surface as cathode At anode: Iron atoms undergo oxidation spontaneously forming $F e^{2+}$ ions$F e \longrightarrow F e^{2+}(a q .)+2 e^{-} \quad E_{o x i}^{o}=-0.44 V$$F e^{2+}$ ions move into solution and electrons into cathodic area where they are picked up by ions of the solution. At cathode:$2 H^{+}+1 / 2 O_{2}+2 e^{-} \longrightarrow H_{2} O E_{\mathrm{red}}^{\circ}=1.23 \mathrm{V}$$\mathrm{H}^{+}$ ions are produced by secondary reactions either from $\mathrm{H}_{2} \mathrm{O}$ or from $\mathrm{H}_{2} \mathrm{CO}_{3}\left(\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}\right)$$H_{2} O \longrightarrow H^{+}+O H^{-}$$H_{2} C O_{3} \longrightarrow H^{+}+H C O_{3}^{-}$The overall reaction of the corrosion cell may be represent$F e(s)+2 H^{+}(a q)+1 / 2 O_{2} \longrightarrow F e^{2+}(a q)+H_{2} O ; E_{c e l l}^{o}=1.67 V$The Fe $\mathrm{Fe}^{2+}$ ions more through water and come at the surface where these are further oxidised into $F e^{3+}$ ions by atmospheric oxygen to form hydrate ferric oxide known as rust, $F e_{2} O_{3} . . H_{2} O$$\begin{aligned} 2 \mathrm{Fe}^{2+}+1 / 2 \mathrm{O}_{2}+2 \mathrm{H}_{2} \mathrm{O} & \longrightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}+4 \mathrm{H}^{+} \\\ \mathrm{Fe}_{2} \mathrm{O}_{3}+\mathrm{xH}_{2} \mathrm{O} \longrightarrow & \mathrm{Fe}_{2} \mathrm{O}_{3} \cdot \mathrm{xH}_{2} \mathrm{O} \end{aligned}$

Q.  The values of $\Lambda_{m}^{\Theta}$ for $N H_{4} C l, N a O H$ and $N a C l$ at infinite are respectively $129.8,248.1$ and 126.4 ohm $^{-1} \mathrm{cm}^{2} \mathrm{mol}^{-1}$ Calculate $\wedge_{m}^{o}$ of $N H_{4} O H$

Q.  Calculate the standard free energy change for the reaction occurring in the cell$Z n(s)\left|Z n^{2+}(1 M) \| C u^{2+}(1 M)\right| C u(s)$$\left[\text { Given } E_{Z n^{2+} / Z n}^{o}=-0.76 V, E_{C u^{2+} / C u}^{o}\right.$$\left.=0.34 V ; F=96500 \mathrm{Cmol}^{-1}\right]$How is it related to equilibrium constant for the reaction

$=0.34 V-(-0.76 V)=1.1 V$Now, $\Delta G^{\circ}=-n F E_{\text {cell }}^{\circ}=-2 m o l \times 96,500 C \mathrm{mol}^{-1} \times 1.10 \mathrm{V}$$=212300 \mathrm{J}=212.300 \mathrm{kJ}$$\Delta \mathrm{G}^{\circ}=2.303 \mathrm{RT} \log k$

Q.  Write Nernst equation and find out the emf of the following cell at $25^{\circ} \mathrm{C}$Pt, Br $_{2} |$ Br $^{-}(0.010 \mathrm{M}) \| \mathrm{H}^{+}(0.030 \mathrm{M}) | \mathrm{H}_{2}(\text { latm }),$ Pt$\left(E_{B r / 2 B r^{-}}^{o}=+1.08 V, \quad R=8.314 J k^{-1} \mathrm{mol}^{-1}, F=96,500\left(\mathrm{mol}^{-1}\right)\right.$

$M^{n+}(a q .)+n e^{-} \longrightarrow M(s)$$E_{M^{n+} / M}=E_{M^{n+} / M}^{o}+\frac{2.303 R T}{n F} \log \left[M^{n+}\right]$where $T$ is temperature, $n$ is valency of the ion, $F$ is faraday constant.Formula $E_{\text {cell }}=E_{r}^{\circ}-E_{l}^{\circ}+\frac{2.303 R T}{n F} \log \left[\frac{H^{+}}{B r^{-}}\right] \quad \ldots$ (i)Here, $E_{c e l l}=? E_{r}^{o}=0, E_{l}^{o}=1.08 \mathrm{V}$$R=8.314, F=96,500 c / \mathrm{mol}, T=298 k$ and $n=1$Substituting these values in the equation (i) we get$E_{\text {cell }}=1.08 \mathrm{V}-0+\frac{2.303 \times 8.314 \times 298}{1 \times 96,500} \log \left[\frac{0.03}{0.01}\right]$$E_{\text {cell }}=1.08 \mathrm{V}+\frac{0.0591}{1} \log 3 \mathrm{V}$$E_{\text {cell}}=1.08+0.0591 \times 0.4771$$E_{\text {cell}}=1.08+0.2819 \mathrm{V}$$=1.36$

Q.  Calculate the potential (emf) of the cell$\mathrm{Cd}\left|\mathrm{Cd}^{2+}(0.10 \mathrm{M}) \| \mathrm{H}^{+}(0.20 \mathrm{M})\right| \mathrm{Pt}, \mathrm{H}_{2}(0.5 \mathrm{atm})$(Given, $E^{o}$ for $C d^{2+} / C d=-0.403 V$ $\left.R=8.314 J k^{-1} \mathrm{mol}^{-1}, F=96,500 \mathrm{Cmol}^{-1}\right)$

Q.  Estimate the minimum potential difference needed to reduce $A l_{2} O_{3}$ at $500^{\circ} \mathrm{C} .$ The face energy change for thedecomposition reaction $\frac{2}{3} A l_{2} O_{2} \longrightarrow \frac{4}{3} A l+O_{2}$ is$\Delta G=+960 \mathrm{kJ},\left(F=96,500 \mathrm{C} \mathrm{mol}^{-1}\right)$

Here $n=6, \Delta G=-n F E$$\therefore 960,000=-6 \times 96500 \times E$$\therefore E=-1.658 V$$\therefore$ Minimum potential difference needed to reduce $A l_{2} O_{3}$ is$1.658 V$

Q.  Calculate the cell emf at $25^{\circ} \mathrm{C}$ for the following cell$M g(s)\left|M g^{2+}(0.01 M) \| S n^{2+}(0.1 M)\right| S n(s)$(Given $E_{M g^{2+} M g}^{o}=-2.34 V . \quad E_{S n^{2}+S n}^{o}=-0.136 V$$I F=96,500 \mathrm{Cmol}^{-1}$Calculate the maximum work that can be accomplished by the operation of this cell

$S n^{2+}(a q .)+2 e^{-} \longrightarrow S n(s)$ (Cathodic half reaction) $M g(s)+S n^{2+}(a q .) \longrightarrow M g^{2+}(a q .)+S n(s)$$E_{c e l l}^{o}=E_{S n^{2+} / S n}^{o}-E_{M g^{2+} / M g}^{o}=-0.136+2.34$\\[ =2.204 V \\]$W_{\max }=-\Delta G^{\circ}=-\left(-n F E_{c e l l}^{o}\right)=n F E_{c e l l}^{o}$$\begin{aligned}=& 2 \times 96,500 \times 2.204 \\\ &=425372 J=425.372 \mathrm{kJ} \end{aligned}$

Q.  (a) Calculate the electrode potential of silver electrode dipped in a 0.1M solution of silver nitrate at $298 K$ assuming silver nitrate to be completely dissociated. The standard electrode potential of $A g^{+} / A g$ is $0.80 V$ at $298 K$(b) At what concentration of silver ions will this electrode have a potential of zero volt

Q.  Depict the galvanic cell in which the reaction$Z n(s)+2 A g^{+}(a q) \rightarrow Z n^{2+}(a q)+2 A g(s)$takes place. Further show:(i) Which of the electrode is negatively charged(ii) The carriers of the current in the cell(iii) Individual reaction at each electrode

$Z n / Z n^{2+}(a q)|| A g^{+}(a g) / A g$(i) In this cell, the $2 n / Z n^{2+}$ will act as negative terminaland $A g / A g^{+}$ will act as positive terminal(ii) The conventional current will flow from silver to zinc electrode in the external circuit(iii) Cathodic reaction: $2 \mathrm{Ag}^{+}+2 e^{-} \longrightarrow 2 \mathrm{Ag}$Anodic reaction: $C u \longrightarrow C u^{2+}+2 e^{-}$

Q.  Write the Nernst Equation and calculate emf. of the following cells at 298 _K_

Thus oxidation will occur at hydrogen electrode and reduction on the $B r_{2}$ electrode. $E_{c e l l}=1.288 \mathrm{V}$

Q.  Predict the products of electrolysis of the following:(i) An aqueous solution of $A g N O_{3}$ with silver electrodes(ii) An aqueous solution of $A g N O_{3}$ with platinum electrodes(iii) A dilute aqueous solution of $H_{2} S O_{4}$ with platinum electrodes(iv) An aqueous solution of cupric chloride with platinum electrodes.(Given $\left.E_{A g^{+} / A g}^{o}=+0.80 V, E_{C u^{2+} / C u}^{o}=+0.34 V\right)$

$A g N O_{3}(s)+a q \rightarrow A g^{+}(a q)+N O_{3}^{-}(a q)$$H_{2} O \rightleftharpoons H^{+}+O H^{-}$At cathode: $A g^{+}$ ions have lower discharge potentialthan $H^{+}$ ions. Hence, $A g^{+}$ ions will be deposited asAg in preference to $H^{+}$ ions.At Anode : As $A g$ anode is attacked by $N O_{3}^{-}$ ions, $A g$of the anode will dissolve to form $A g^{+}$ ions in the solution.$A g \longrightarrow A g^{+}+e^{-}$(ii) Electrolysis of aqueous solution of $A g N O_{3}$ using platinum electrodes. At Cathode : Same as above.At Anode : As anode is not attackable, out of $\mathrm{OH}^{-}$ and$\mathrm{NO}_{3}^{-}$ ions, OH^- ions have lower discharge potential. Hence, $\mathrm{OH}^{-}$ ions will be discharged in preference to$\mathrm{NO}_{3}^{-}$ ions, which then decompose to given out $\mathrm{O}_{2}$$O H^{-}(a q) \longrightarrow O H+e^{-}, 4 O H \rightarrow 2 H_{2} O(l)+O_{2}(g)$(iii) Electrolysis of dilute $H_{2} \mathrm{SO}_{4}$ with platinum electrodes.$H_{2} S O_{4}(a q) \rightarrow 2 H^{+}(a q)+S O_{4}^{2-}(a q)$\\[ \begin{array}{l} {H_{2} O \rightleftharpoons H^{+}+O H^{-}} \\\ {\text {At Cathode: } H^{+}+e \rightarrow H, \quad H+H \rightarrow H_{2}(g)} \end{array} \\]At Anode: $O H^{-} \rightarrow O H+e, \quad 4 O H \rightarrow 2 H_{2} O+O_{2}(g)$Thus, $\mathrm{H}_{2}$ is liberated at the cathode and $\mathrm{O}_{2}$ at the anode.(iv) Electrolysis of aqueous solution of $\mathrm{CuCl}_{2}$ with platinum electrodes$\mathrm{CuCl}_{2}(\mathrm{s})+a q \rightarrow \mathrm{Cu}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q)$$H_{2} O \rightleftharpoons H^{+}+O H^{-}$At Cathode: $\mathrm{Cu}^{2+}$ ions will be reduced in preferenceto $\mathrm{OH}^{-}$ ions $\mathrm{Cl}^{-} \rightarrow \mathrm{Cl}+e^{-}, \mathrm{Cl}+\mathrm{Cl} \rightarrow \mathrm{Cl}_{2}(\mathrm{g})$Thus, $C u$ will be deposited on the cathode and $C l_{2}$ will eb liberated at the anode.

Q.  Predict the products of electrolysis of each of the following:(i) An aqueous solution of A g N O_{3} using silver electrodes.(ii) An aqueous solution of using platinum electrodes(iii) A dilute solution of using platinum electrodes.(iv) An aqueous solution of using platinum electrodes

At cathode : $A g^{+}$ ions with less discharge potential are reduced in preference to $\mathrm{H}^{+}$ ions which will remain in solution. As a result, silver will be deposited at cathode.$A g^{+}(a q)+e^{-} \rightarrow A g \quad(\text { deposited })$At anode : An equivalent amount of silver will be oxidised to Ag” ions by releasing electrons.$A g(s)-\stackrel{(a q)}{\longrightarrow} A g^{+}(a q)+e^{-}$Thus, the concentration of the solution will remain unaffected.(ii) An aqueous solution of $A g N O_{3}$ using platinum electrodes: In the case the platinum electrodes are the non-attackable electrodes. On passing current the following changes will occur at the electrodes. At cathode : Ag” ions will be reduced to Ag which will get deposited at the cathode.At anode : Both $\mathrm{NO}_{3}^{-}$ and $\mathrm{OH}^{-}$ ions will migrate. But $\mathrm{OH}^{-}$ ions with less discharge potential will be oxidise in preference to $\mathrm{NO}_{3}^{-}$ ions which will remain in solution.$\mathrm{OH}^{-}(a q) \rightarrow \mathrm{O}+e^{-} ; 4 \mathrm{OH} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)$Thus, as a result of electrolysis, silver is deposited on the cathode while $\mathrm{O}_{2}$ is evolved at the anode. The solution will be acidic due to the presence of $H N O_{3}$(iii) A dilute solution of $H_{2} \mathrm{SO}_{4}$ using platinum electrodes: On passing current, both acid and water will ionise as follows:$H_{2} S O_{4}(l)-\stackrel{(a q)}{\longrightarrow} 2 H^{+}(a q)+S O_{4}^{2-}(a q)$At cathode : $\mathrm{H}^{+}(\mathrm{aq})$ ions will migrate to the cathode and will be reduced to $\mathrm{H}_{2}$H^{+}(a q)+e^{-} \longrightarrow H ; H+H \longrightarrow H_{2}(g)At anode : $O H^{-}$ ions will be released in preference to $\mathrm{SO}_{4}^{2-}$ ions because their discharge potential is less.They will be oxidised as follows:$O H^{-}(a q) \rightarrow O H+e^{-} ; 4 O H \rightarrow 2 H_{2} O(l)+O_{2}(g)$Thus, $\mathrm{O}_{2}(g)$ will be evolved at anode. The solution will be acidic and will contain $H_{2} S O_{4} .$(iv) An aqueous solution of $\mathrm{CuCl}_{2}$ using platinum electrodes:The electrolysis proceeds in the same manner as discussed in the case of $\mathrm{AgNO}_{3}$ solution.Both $\mathrm{CuCl}_{2}$ and $\mathrm{H}_{2} \mathrm{O}$ will ionise as follows:\begin{array}{l}{\mathrm{CuCl}_{2} \frac{(\mathrm{aq})}{2}, \mathrm{Cu}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q)} \\\{\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q)} \end{array}At cathode: $\mathrm{Cu}^{2+}$ ions will be reduced in preference to $\mathrm{H}^{+}$ ions and copper will be deposited at cathode$C u^{2+}(a q)+2 e^{-} \longrightarrow C u$ (deposited)At anode: $\mathrm{Cl}^{-}$ ions will be discharged in preference to $\mathrm{OH}^{-}$ ions which will remain in solution.$C l^{-} \rightarrow C l^{-}+e^{-} ; C l+C l \rightarrow C l_{2}(g)$Thus, Cl, will be evolved at anode.

Q.  (a) Calculate the emf of the cell(b) Explain with example the terms weak and strong electrolytes**OR**(a) The resistance of a conductivity cell containing $0.001 M K C l$ solution at $298 K$ is $1500 \Omega$ what is the cell constant, it the conductivity of $0.001 \mathrm{MKCl}$ solution at $298 K$ is $0.146 \times 10^{-3} \mathrm{S} \mathrm{cm}^{-1}$(b) Predict the product of electrolysis in the solution of H_{2} S O_{4} with platinum electrode

$M g-2 e^{-} \longrightarrow M g^{2+} ; E^{\circ}=+2.37 V$$E_{\text {cell }}^{\circ}=E_{M g / M g^{2+}}^{\circ}+E_{C u^{2+} / C u}^{\circ}$$=(2.37+0.34) V=3.71 V$(b) Strong electrolytes are almost completely ionized when dissolved in water or any other polar solvent. In solution they are excellent conductors e.g. $N a O H$$N a C l, K N O_{3}, H C l$ etc.Weak electrolytes are feebly ionized when dissolved in water or any other polar solvent. they are bad conductors of electricity e.g., $\mathrm{CH}_{3} \mathrm{COOH}, \mathrm{H}_{3} \mathrm{PO}_{4}, \mathrm{NH}_{4} \mathrm{OH}$ etc.OR(a) $\quad k=\frac{1}{R} \cdot \frac{l}{a}$where $k=$ conductivity, $R=$ resistance, $l / a=\mathrm{cell}$constant or $1 / a=R . K=1500 \times 0.146 \times 10^{-3} \mathrm{cm}$ $=0.219 \mathrm{cm}$For dilute solution of $H_{2} \mathrm{SO}_{4}$ reaction (i) is preferred;in that case $\mathrm{O}_{2}$ is liberated at anode.But at higher concentration of $H_{2} S O_{4}$ reaction (ii) ispreferred and peroxodisulphuric acid $\left(H_{2} S_{2} O_{8}\right)$ solution is obtained.

Q.  Calculate the emf of the cell$Z n / Z n^{2+}(0.1 M) \| C d^{2+}(0.01 M) | C d$ at $298 k$[Given $E_{z n^{2+} / Z n}^{o}=-0.76 V$ and $E_{C d^{2}+C d}^{o}=-0.40 V$

Q.  Calculate the emf of the cell$Z n / Z n^{2+}(0.1 M) \| C d^{2+}(0.01 M) | C d$ at $298 k$[Given $E_{z_{n} n_{Z n}}^{o}=-0.76 V$ and $E_{C d^{2}+C d}^{o}=-0.40 \mathrm{V}$OR(a) Account for the following(i) Alkaline medium inhibits the rusting of iron(ii) Iron does not rust even if the Zinc coating is broken in a galvanized iron pipe.(i) Construct a galvanic cell using the above data(ii) For what concentration of $A g^{+}$ ions will the emf of the cell be zero at $25^{\circ} \mathrm{C}$. If the concentration of$C u^{2+}$ is $0.01 M ? \quad[\log 3.919=0.593]$

(a) (i) The alkaline medium prevents the availability of $H^{+}$ ions. This in turn reduces rate of oxidation of $F e$ to $\mathrm{Fe}^{2+} .$ Thus the rusting of iron is inhibited.(ii) Even if the zinc coating is broken in an iron pipe, the remaining zinc layer undergoes oxidation in the remaining zinc layer undergoes oxidation in preference to iron because of its more electropositive nature than iron. Hence iron does not rust.\text { (b) } \quad \text { (i) Since } E_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{\circ} \text { is greater than } E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ} \mathrm{SO} reduction will occur at silver electrode. The cell representation is $C u(s)\left|C u^{2+}\right|\left|A g^{+}\right| A g(s)$

Q.  When a certain conductivity cell was filled with $0.1 \mathrm{MKCl}$, it has a resistance of $85 \mathrm{W}$ at $25^{\circ} \mathrm{C}$. When the same cell was filled with an aqueous solution of $0.052 M$ unknown electrolyte, the resistance was $96 \mathrm{W}$. Calculate the molar conductivity of the unknown electrolyte at this concentration. (Specific conductivity of $0.1 M K C l=1.29 \times 10^{-2}$ ohm $^{-1} \mathrm{cm}^{-1}$ )

Specific conductance, $k=1.29 \times 10^{-2} \Omega^{-1} \mathrm{cm}^{-1}$Cell constant $=k \times R=1.29 \times 10^{-2} \Omega^{-1} \mathrm{cm}^{-1} \times 85 \Omega=1.1 \mathrm{cm}^{-1}$Resistance of unknown electrolytic solution, $R=96 \Omega$Specific conductance, $k=\frac{\text { Cell constant }}{R}=\frac{1.1 \mathrm{Cm}^{-1}}{96 \Omega}$\\[ =\frac{11}{960} \Omega^{-1} \mathrm{cm}^{-1} \\]\begin{aligned}&\text { Concentration, } C=0.052 \frac{\mathrm{mol}}{L}=0.052 \frac{\mathrm{mol}}{1000 \mathrm{Cm}^{3}}\\\&=5.2 \times 10^{-5} \mathrm{mol} \mathrm{cm}^{-3}\\\&\text { Molar coductance, } \wedge_{m}=\frac{11 \Omega^{-1} C m^{-1}}{960 \times 5.2 \times 10^{-5} \mathrm{molcm}^{-3}}\\\&=220.2 \mathrm{\Omega}^{-1} \mathrm{Cm}^{2} \mathrm{mol}^{-1}\end{aligned}

Q.  Calculate the potential of the following cell reaction at 298K.$S n^{4+}(1.50 M)+Z n(S) \longrightarrow S n^{2+}(0.50 M)+Z n^{2+}(2.0 M)$The standard potential $E^{o}$ of the cell is $0.89 V$ whether the potential of the cell will increase or decrease, if the concentration of $S n^{4+}$ is increased in the cell? $\left(R=8.314 J K^{-1} \mathrm{mol}^{-1} ; F=96,500 \mathrm{Cmol}^{-1}\right)$

Q.  Calculate the cell emf and $G$ for the cell reaction at $25^{\circ} \mathrm{C}$ for the cell

The net cell reaction is $Z n(s)+C d^{2+}(a q .) \rightarrow Z n^{2+}(a q .)+C d(s)$Here value of $n=2$$\therefore \Delta G^{\circ}=-n F E_{c e l l}^{o}=-2 \times 96,500 \times 0.36=69480 J$

Q.  Using the standard electrode potentials given below, predict if the reaction between the following is feasible:

In this reaction $F e^{3+}$ ions are reduced by $I^{-}$ ions. For this reaction to be feasibly the $E_{F e^{3+} / F e^{2+}}^{\Theta}$ should be largerthan $E_{I_{2} / I^{-}}^{\Theta} .$ since the given values agree with this requirement. Hence this reaction is Feasible.\text { (b) } \quad 2 A g=(a q)+C u(s) \longrightarrow C u^{2}+2 A gIn this reaction, $A g^{+}$ ions are being reduced. This isalso feasible because $E_{F e^{3+} / F e^{2+}}^{\Theta}$ is less than $E_{B_{12} / B r}^{\Theta} .$ Hence Br ions cannot reduce Fe’t ions.

Q.  Predict the products of electrolysis in each of the following:(i) An aqueous solution of $A g N O_{3}$ with silver electrodes(ii) An aqueous solution of $A g N O_{3}$ with platinum electrodes(iii) A dilute solution of $H_{2} S O_{4}$ with platinum electrodes(iv) An aqueous solution of $\mathrm{CuCl}_{2}$ with platinum electrodes.. \begin{aligned}Given that $E_{r e d}^{\Theta} H_{2} O=-0.41 V ; \quad E_{o x i}^{\theta}$ of$H_{2} O=-1.23 V ; E_{A g^{+} / A g}^{\theta}=+0.80 V ; \quad E_{C u^{2+} / C u}^{\theta}=0.34 V$

At cathode : $A g^{+}$ ion will reduce in preference to $H_{2} O$$A g^{+}+e^{-} \longrightarrow A g$At anode: $A g$ will get oxidise in preference to $\mathrm{H}_{2} \mathrm{O}$$A g \longrightarrow A g^{+}+e^{-}$(ii) $\quad A g N O_{3} \longrightarrow A g^{+}+N O_{3}^{-}$At cathode: $\quad A g^{+}+e^{-} \longrightarrow A g$At anode: $\mathrm{H}_{2} \mathrm{O}$ will oxidise in preference to $\mathrm{NO}_{3}^{-}$ ions$H_{2} O \longrightarrow 2 H^{+}+\frac{1}{2} O_{2}+2 e^{-}$\begin{aligned} &\text { (iii) } \quad H_{2} S O_{4} \longrightarrow 2 H^{+}+S O_{4}^{2-}\\\ &\text { At cathode : } 2 H^{+}+2 e^{-} \longrightarrow H_{2}\\\ &\text { At anode : } H_{2} O \longrightarrow 2 H^{+}+\frac{1}{2} O_{2}+2 e^{-} \end{aligned}(iv) $\quad \mathrm{CuCl}_{2} \longrightarrow \mathrm{Cu}^{2+}+2 \mathrm{Cl}^{-}$At cathode : $\mathrm{Cu}^{2+}$ will get reduced in preference to $\mathrm{H}_{2} \mathrm{O}$$\mathrm{Cu}^{2+}+2 e^{-} \longrightarrow \mathrm{Cu}$At anode: $\mathrm{Cl}^{-}$ ion will oxidise in preference to $\mathrm{H}_{2} \mathrm{O}$$2 \mathrm{Cl} \longrightarrow \mathrm{Cl}_{2}+2 \mathrm{e}^{-}$