Question

A solution containing 1.0 M NiCl2 and 1.0 M SnBr2 undergoes electrolysis by passing a current between two platinum electrodes. What are the most likely first products to be formed? Cl2 (aq) + 2e- → 2Cl- (aq) ℰ°= 1.36 V Br2 (aq) + 2e- → 2Br- (aq) ℰ° =1.08 V Sn2+ (aq) + 2e- → Sn(s) ℰ° = -0.14 V Ni2+ (aq) + 2e- → Ni(s) ℰ°= -0.24 V

Answer 1

Answer:

Sn2+ (aq) + 2e- → Sn(s) ℰ° = -0.14 V

Explanation:

A close look at all the options shows that the most feasible first reaction is the reduction of tin II ion to ordinary metallic tin.

Given the two half cells, nickel is oxidized in one half cell to Ni II while in the second half cell, tin II ion is reduced to metallic tin. The platinum electrodes simply act as electron conduits in the cell.