In this electrolysis experiment, we're asked to determine how many grams of lead will be deposited from a solution containing lead to ions. The current is 0.995 and is applied for 30.9 minutes. The first thing we need to do is calculate the total charge in cool-arms using the current and the time in seconds. The current is in the units of cool-arms per second. So we multiply the cool-arms per second by the total amount of seconds in the 30.9 minutes. And we end up getting 1771 cool-arms. The next thing we do is we calculate the moles of electrons using ferrities constant. And that gives us 0.0183 moles of electrons. The point 0138 moles is of electrons not of lead. That's a real important distinction you want to make when you're doing these kinds of problems. The third step is to calculate the moles deposited using the mole ratio of electrons to the metal formed. So let's look at the half reaction for the reduction of lead 2 to lead solid. For every one mole of lead ion, two moles of electrons are required. So therefore, we're going to use the point 0183 moles as follows. We're going to divide that mole amount by 2 to get the moles of lead. And finally, we convert the moles of lead deposited to grams of lead. Here we are asked to determine the grams of silver that will be deposited from a silver solution with a current of 1.21 amps or 1.21 cool-arms per second. That is applied for 31.1 minutes. First, let's calculate the total charge in cool-arms using the current cool-arms per second and the time. We end up with 1,782 cool-arms. Second, let's calculate the moles of electrons using ferrities constant. 0.0185 moles of electrons. Third, calculate the moles of silver deposited using the mole ratio of electrons to the metal formed. Here in the reduction of silver ion, one electron is required. So it's a 1 to 1 ratio between moles of electrons and moles of silver deposited. And finally, we'll calculate the grams of silver deposited using the molar mass of silver. Rather than determining the amount of metal deposited, here we are asked to determine the time required to deposit a certain amount of metal. How many seconds are required to deposit 0.285 grams of cadmium using a current of 0.956 amps? First, calculate the moles of cadmium deposited, converting the grams of cadmium to moles of cadmium. Second, we're going to determine the moles of electrons required for this reduction. So we need to take a look at the cadmium 2 plus reduction half-reaction. We see that two moles of a required for every mole of cadmium ion. So moles of cadmium metal, same as moles of cadmium ion. So we convert the moles of cadmium metal to moles of electrons considering the 2 to 1 ratio in this half-reaction. Now, we're going to convert the moles of electrons to coulombs using faredness constant. And finally, we'll calculate the time by dividing the coulombs by the inverse of the current where the reciprocal of the current. How many seconds are required to deposit 0.285 grams of chromium from a solution that contains chromium 3 plus ions using a current of 1.06 amps? First, convert the grams of chromium by dividing it by the molar mass of chromium. Second, calculate the number of moles of electrons by using the mole ratio of electrons to the metal. Here, the reduction half-reaction for chromium 3 plus to chromium solid requires 3 electrons for every 1 mole of chromium ion. We use that ratio and calculate 0.0162 moles of electrons. Then, we calculate the total charge and coulombs using faredness constant. And finally, calculate the time in seconds by multiplying the coulombs by the reciprocal of the current, who are essentially dividing the coulombs by the value of the current.