Consider an electrochemical cell based on the following cell diagram:

Question

Pt | Pu3+(aq), Pu4+(aq) || Cl2(g), Cl−(aq) | Pt

Given that the standard cell emf is 0.35 V and that the standard reduction potential of chlorine is 1.36 V, what is the standard reduction potential E°(Pu4+/Pu3+), in volts? [Enter only a number.]

Answer 1

Pu+3 ==> Pu+4 + e Eo = ??
Cl2 + 2e ==> 2Cl^- Eo 1.36
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2Pu+3 + Cl2 ==> 2Pu+4 + 2Cl^- Ece.. = 0.35
What is ?? + 1.36 = 0.35
Solve for ?? which will be the oxidation potential (because that's the way the half cell is written above) so change the sign for the reduction problem (what the problem is asking).

Answer 2

Cl is gaining elections which means it is the reduction half reaction. Reduction is associated with the Cathode.

E(cell) = E(cathode) - E(anode)

0.35v = 1.36v - E(anode)
solving for E(anode) gives you 1.01v

Answer 3

1.17 V