Question

A cell is constructed by dipping a zinc rod in 0.1 M zinc nitrate solution and a lead rod in 0.2 M lead nitrate solution.

\[\ce{E^{\circ}_{Pb^{2+}/Pb}}\] = −0.13 V and \[\ce{E^{\circ}_{Zn^{2+}/Zn}}\] = −0.76 V

1. Write the spontaneous cell reaction.
2. Calculate standard emf and emf of the cell.

Answer 1

Given: \[\ce{Zn_{(s)} ∣ Zn{^{2+}} (0.1M) ∣∣ Pb^{2+} (0.2M) ∣ Pb_{(s)}}\]

\[\ce{E^{\circ}_{{pb^{2+}/{Pb}}}}\] = −0.13 V and

\[\ce{E^{\circ}_{{Zn^{2+}/{Zn}}}}\] = −0.76 V

Spontaneous cell reaction: \[\ce{Zn_{(s)} + Pb^{2+}_{ (aq)} −> Zn^{2+}_{ (aq)} + Pb_{(s)}}\]

\[\ce{E{^{\circ}_{cell}} = E{^{\circ}_{cathode}} - E{^{\circ}_{anode}}}\]

= (−0.13V) − (−0.76V)

= +0.63 V

Using the Nernst equation:

\[\ce{E = E^{\circ}_{cell} - \frac{0.0591}{n} log \frac{[Zn^{2+}]}{[Pb^{2+}]}}\]

n = 2 electrons exchanged

[Zn²⁺] = 0.1 M, [Pb²⁺] = 0.2 M

\[\ce{E = 0.63 - \frac{0.0591}{n} log \frac{0.1}{0.2}}\]

= 0.63 − 0.02955 log (0.5)

= 0.63 − 0.02955 × (−0.3010)    ...(log 0.5 = −0.3010)

= 0.63 + 0.0089

= 0.6389 V

≈ 0.639 V