Question

An electrochemical cell is powered by the half reactions shown below.

Answer 1

Ag+(aq) + e– → Ag(s)     Eo = +0.80 V
Pb2+(aq) + 2e– → Pb(s) Eo = –0.13 V

=>

2Ag+(aq) + 2e– → 2Ag(s)     Eo = +0.80 V

Pb(s) → Pb2+(aq) + 2e–      - Eo = – (-0.13 V) = +0.13 V
----------------------------------

2Ag(+) (aq) + Pb(s) -------> 2Ag(s) + Pb(2+)


E cell = E0 cathode – Eo anode = 0.80V - (- 0.13V) = 0.80V + 0.13V = 0.93V.

Answer: + 0.93V