assigning half reactions to oxidation/reduction

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

Erin 2I
Posts: 41
Joined: Fri Sep 25, 2015 3:00 am

assigning half reactions to oxidation/reduction

For number 14.37, we need to determine the potential of the following cell:

Pt(s) | Cl2(g) | HCl(aq) || HCl(aq) | H2(g) | Pt(s)

The two half reactions are
Cl2 + 2e- --> 2C- E˚ = +1.36
2H+ + 2e- --> H2 E˚ = 0

The book says that the cathode will be H reaction, and the anode will be the Cl reaction. First of all,how do we determine which reaction will be the oxidation, and which will be the reduction? I thought that the H reaction should be the anode, because the E˚ is more "more negative", or just plain smaller than the E˚ for Cl, meaning that it was the reducing agent, meaning that it is being oxidized, meaning that it is the anode. Also, if we made H reaction the anode, then the E˚ of the cell will be positive, which is what we want. I am very confused.

Nathan Danielsen 1G
Posts: 17
Joined: Fri Sep 25, 2015 3:00 am

Re: assigning half reactions to oxidation/reduction

When given the cell diagram for a reaction, the anode and oxidation reaction are always on the left, and the cathode and reduction reaction are always on the right. In part C of 14.37, you are given the H+ reaction on the right, and must treat it as the cathode. Even though the final answer means the cell isn't favorable, that does not matter in the context of the problem. We just need to find E, regardless of whether or not it is negative or positive.

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