## 6O.1 and 6O.3

$E_{cell} = E_{cell}^{\circ}-\frac{RT}{nF}\ln Q$

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### 6O.1 and 6O.3

For problem 6O.1 we chose the cathode to be the nickel reaction which had a smaller or more negative E and this makes sense because the reaction is electrolyzed which means that a non spontaneous reaction was driven by an outside current allowing for the cathode to have a smaller E than that of the anode.
However, for problem 6O.3, according to the answers, we are choosing the more positive or bigger number to be the cathode which implies that the reaction is spontaneous. Are we supposed to assume that the question is implying that these reactions are spontaneous even though it says all of the concentrations are electrolyzed? Thanks.

Ryan Narisma 4G
Posts: 104
Joined: Fri Aug 30, 2019 12:18 am

### Re: 6O.1 and 6O.3

Hi! To answer your question, in an electrolytic cell you are trying to precipitate out metal solids from their ionic constituents through input of electrical energy in the solution. In this case we have Nickel (II) Sulfate, so the end goal is to reduce the nickel ion into nickel solid. Because the metal's ions are in solution, we must take into account of water. Water can be both oxidized and reduced, so you have to choose the oxidation reaction and reduction reaction which would result in the least negative voltage. To go about the problem, I like to look at what are the potential reactants in the reaction: Ni2+, H2O, or SO42-. In this problem we don't want to look at the reduction of sulfate because we want the precipitate out the nickel. So we know that nickel can be reduced, but water can be oxidized or reduced. Comparing the cell potential at standard conditions for each of them, it can be gathered that nickel has a more positive voltage than water when reduced. But, in a redox reaction, we need something oxidized and something reduced. The only reactant that can be oxidized is water, so we have no choice but to include water in our oxidation reaction. Combining the two half reactions yields a negative voltage of -1.46V, which is expected as this is an electrolytic cell.

Similarly, in question O.3, you are doing the exact same thing: write out the reduction reactions and all of the oxidation reactions and pick and choose which two would give you the cell with a less negative voltage. I hope this helps!

Lindsey Chheng 1E
Posts: 110
Joined: Fri Aug 30, 2019 12:16 am

### Re: 6O.1 and 6O.3

Ryan Narisma 4G wrote:Hi! To answer your question, in an electrolytic cell you are trying to precipitate out metal solids from their ionic constituents through input of electrical energy in the solution. In this case we have Nickel (II) Sulfate, so the end goal is to reduce the nickel ion into nickel solid. Because the metal's ions are in solution, we must take into account of water. Water can be both oxidized and reduced, so you have to choose the oxidation reaction and reduction reaction which would result in the least negative voltage. To go about the problem, I like to look at what are the potential reactants in the reaction: Ni2+, H2O, or SO42-. In this problem we don't want to look at the reduction of sulfate because we want the precipitate out the nickel. So we know that nickel can be reduced, but water can be oxidized or reduced. Comparing the cell potential at standard conditions for each of them, it can be gathered that nickel has a more positive voltage than water when reduced. But, in a redox reaction, we need something oxidized and something reduced. The only reactant that can be oxidized is water, so we have no choice but to include water in our oxidation reaction. Combining the two half reactions yields a negative voltage of -1.46V, which is expected as this is an electrolytic cell.

Similarly, in question O.3, you are doing the exact same thing: write out the reduction reactions and all of the oxidation reactions and pick and choose which two would give you the cell with a less negative voltage. I hope this helps!

Hi Ryan! Thank you for such a clear explanation! I did what you said to do for 6O.3, but for part c, I got that H2O is reduced and not Ni2+. I calculated that having H2O at the cathode would make the overall voltage more positive (+1.05V) than if Ni2+ was at the cathode (-1.05V). Do you know why the answer is Ni2+?

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