General Chemistry

Question

1. Draw cell diagrams of galvanic cells one of which Zn is the anode, the other − the cathode. Write the electrode reactions. Calculate E0cell.

 

2. Calculate the cell potential of Mg/Nigalvanic cell when the ion concentrations are: C(Mg2+)= 0.1 M; C(Ni2+) = 1 M. Draw cell diagram of galvanic cell and write the electrode reactions. 

Answer 1

Anode on the left, cathode on the right

We'll use Cu as the cathode

Zn | Zn²⁺(aq) || Cu²⁺ | Cu

Electrode reactions:

Zn ==> Zn²⁺ + 2e- oxidation Eº = -0.76V

Cu²⁺ + 2e- ==> Cu reduction Eº = +0.34

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Zn(s) + Cu²⁺(aq) ==> Zn²⁺(aq) + Cu(s) overall redox equation

Eº_cell = 0.34 + 0.76 = 1.10V

Cell diagram:

Mg | Mg²⁺(0.1M) || Ni²⁺(1 M) | Ni

Electrode reactions:

Mg ==> Mg²⁺ + 2e- oxidation Eº = -2.38V

Ni²⁺ + 2e- ==> Ni reduction Eº = -0.23V

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Mg(s) + Ni²⁺(aq) ==> Mg²⁺⁽aq) + Ni(s) overall redox equation

Eº_cell = -0.23 -(-2.38) = +2.15V

To calculate the E_cell, we use the Nernst equation:

E_cell = Eº_cell - 0.0592/n log Q ... assuming temperature = 298K

Q = [Mg²⁺]/[Ni²⁺] = 0.1/1 = 0.1

E_cell = 2.15 - 0.0592/2 log 0.1

E_cell = 2.15 - (0.0296)(-1) = 2.15 + 0.0296

E_cell = 2.18V