Question

A voltaic cell consisting of Fe²⁺(aq)| Fe(s) and Bi³⁺(aq)| Bi(s) electrodes is constructed. When the circuit is closed, mass of Fe electrode decreases and that of Bi electrode increases.

1. Write cell formula .
2. Which electrode is cathode and which electrode is anode?
3. Write electrode reactions and overall cell reaction.

Answer 1

1. Since the mass of Fe electrode decreases, it undergoes oxidation and it is an anode or an oxidation electrode while as the mass of Bi electrode increases, there is a reduction of Bi³⁺ to Bi and it is cathode or a reduction electrode. Hence the cell formula is,
   \[\ce{Fe(s) | \underset{(1 M)}{Fe^2+}(aq) | \underset{(1 M)}{Bi^3+}(aq) | Bi}\]
2. The left hand electrode is an anode and right hand electrode is a cathode.
3. Reactions:
   LHE (Fe(s) → Fe²⁺(aq) + 2e⁻) × 3     ....(Oxidation half reaction)
   RHE (Bi³⁺(aq) + 3e⁻ → Bi(s)) × 2      ....(Reduction half reaction)
   Overall reaction:
   3Fe(s) + 2Bi³⁺(aq) → 3Fe²⁺ +2Bi(s)