Electrochemistry

Question

A zinc electrode is submerged in an acidic 0.80M Zn+ solution which are connected by salt bridge to a 1.30M Ag+ solution containing a silver electrode. Determine the initial voltage of the cell at 298k

Given that:

Zn/Zn2+ E°=0.76V

Ag+/Ag E°=0.80V

Answer 1

Zn2+ + 2e- ==> Zn Eº = 0.76 V (anode)

Ag+ + e- ==> Ag Eº = 0.80 V (cathode)

Overall reaction:

2Ag⁺ + Zn ==> 2Ag(s) + Zn²⁺

Eº = cathode - anode = 0.80 - 0.76 = 0.04 V

Using the Nernst equation to find Ecell under the non-standard conditions:

Ecell = Eº - RT / nF ln Q

At 298K, the Nernst reduces to

Ecell = Eº - 0.0592 / n log Q

n = 2 mols electrons

Q = [Zn²⁺] / [Ag⁺]² = (0.80) / (1.30)² = 0.473

Ecell = 0.04 - 0.0296 log 0.473 = 0.04 + 0.0096

Ecell = 0.0496 V

(be sure to check the math)