Question

The e.m.f and the standard e.m.f of a cell in the following reaction is 5 V and 5.06 V at room temperature, Ni(s) + 2Ag^+(n) → Ni^2+(0.02M) + 2Ag(s). What is the concentration of Ag^+ ions?

(a) 0.0125 M

(b) 0.0314 M

(c) 0.0625 M

(d) 0.0174 M

I had been asked this question by my college director while I was bunking the class.

My question is taken from Electrochemistry topic in section Electrochemistry of Chemistry – Class 12

Answer 1

Correct option is (d) 0.0174 M

The best I can explain: Given, Temperature T = 298K

Concentration of Ni^2+ = (0.02M)

E(cell) = E°(cell) – \(\frac{0.059}{n}\)log10 (Anode / Cathode)

5 = 5.06 – \(\frac{0.059}{2}\)log10(0.02 / [Ag^+]^2)

[Ag^+]^2 = 0.0174 M.