A voltaic cell contains one half-cell with zinc electrode in a Zn^2+(aq) solution and a copper electrode in a Cu2+(aq) solution.

Question

A voltaic cell contains one half-cell with zinc electrode in a Zn²⁺(aq) solution and a copper electrode in a Cu²⁺(aq) solution. At standard conditions, E° = 1.10 V. Which condition below would cause the cell potential to be greater than 1.10 V?

 

0.1 M Zn²⁺(aq), 1.0 M Cu²⁺(aq)

5.0 M Zn²⁺(aq), 5.0 M Cu²⁺(aq)

5.0 M Zn²⁺(aq), 1.0 M Cu²⁺(aq)

1.0 M Zn²⁺(aq), 1.0 M Cu²⁺(aq)

0.5 M Zn²⁺(aq), 0.5 M Cu²⁺(aq)

Answer 1

Look at the redox reaction taking place:

Copper is the cathode and Zn is the anode (Cu has a greater reduction potential)

Cu²⁺ + Zn ==> Cu + Zn²⁺ Eºcell = 0.34 + 0.76 = 1.10 V

From the Nernst equation we see that Ecell is related to the concentrations as follows:

Ecell = Eºcell - RT/nF ln Q and under standard conditions, this reduces to...

Ecell = Eºcell - 0.0592/n log Q

For Ecell to be greater than Eºcell, the value of log Q must be negative so Q must be LESS THAN 1

Q is the ratio of Zn²⁺ / Cu²⁺

So, 0.1 M Zn²⁺(aq), 1.0 M Cu²⁺(aq) would be the appropriate answer.