Question
Write the Nernst equation and emf of the following cells at 298 k.
Sn(s) | Sn2+ (0.050 M) || H+(0.020 M) H2(g) (1bar) Pt (s)
Solution
For the given cell Anode reaction:
Sn(s) → Sn2+ (aq) + 2e–
Cathode reaction:
2H+(aq) + 2e– → H2(g)
Overall cell reaction:
Sn(s) + 2H+ (aq) → Sn2+(aq) + H2(g)
Here, n = 2, E°cell = E°cathode – E°anode = 0 – (– 0.14 V) = + 0.14 V.
The Nernst equation for Ecell and 298 k can be written as:
