Number 5B from 2015 Final

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Chris_Rudewicz_3H
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Number 5B from 2015 Final

Postby Chris_Rudewicz_3H » Sat Mar 18, 2017 9:37 pm

"A standard electrochemical cell is made by placing a silver electrode into a 1.0 M Ag+ solution and a cadmium electrode into a 1.0 M Cd2+ solution. What is the redox reaction and what is the maximum potential, Ecell, produced by this cell?"

How do I know which one is being oxidized and which one is being reduced?

The answers say the Cadmium (Cd) is being oxidized and therefore is the anode.

nickrodgers
Posts: 10
Joined: Fri Jul 22, 2016 3:00 am

Re: Number 5B from 2015 Final

Postby nickrodgers » Sat Mar 18, 2017 10:18 pm

I assume we'll be given the electric potentials of the relevant half reactions, or else I don't see how it's possible to do this.

In that case, the one with the more positive electric potential is the stronger reducing agent and the one with the more negative electric potential is the oxidizing agent.

Caity Colt 2N
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Joined: Wed Sep 21, 2016 2:59 pm

Re: Number 5B from 2015 Final

Postby Caity Colt 2N » Sat Mar 18, 2017 10:20 pm

You want to find the maximum cell potential, you want the largest, positive E˚cell possible. Since:
Cd2+ + 2e- --> Cd(s) E˚= -0.40V
Ag+ + 1e= --> Ag(s) E˚= +0.80V

Flipping the Cd reaction would give the largest E˚cell since you would reverse the sign to E˚= +0.40, it would be the anode since the Cd would be oxidized (loses electrons).
So:
Cd(s) --> Cd2+ + 2e- E˚= +0.40V
Ag+ + 1e= --> Ag(s) E˚= +0.80V

E˚cell= 1.20V


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