## Test 2 Question 8

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

Paula Sing 1J
Posts: 36
Joined: Fri Sep 29, 2017 7:06 am

### Test 2 Question 8

For part c of this problem when you calculate the E(cell) value, I know it's usually cathode minus anode, but that is when you are working with two reduction potential. In the problem you are given an oxidation potential and a reduction potential, so I added the two together (0.22+ (-)0.15)= 0.07 but I realize this isn't right. How are you supposed to solve this?

Cynthia Tsang
Posts: 52
Joined: Fri Sep 29, 2017 7:07 am
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### Re: Test 2 Question 8

There are two ways to calculate E(cell) with the E(knot) values of the half reactions.

Method 1:
1. Reverse the anode half-reaction.
2. Keep the E(knot) value the same.
3. Use E(cathode) - E(anode) without changing the sign for the anode reaction, but using the reverse reaction.

Method 2:
1. Reverse the anode half-reaction.
2. Multiply the E(knot) value by -1.
3. Add the E(knot) values of the half reactions together.

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