Question

Calculate:

E_cell at 25°C for the reaction:

\[\ce{Zn + Cu{^{2+}} (0.20 M) -> Zn{^{2+}} (0.50 M) + Cu}\]

Given: \[\ce{E^{\circ}_{({Zn^{2+}/{Zn}})}}\] = −0.76 volt, \[\ce{E^{\circ}_{({Cu^{2+}/{Cu}})}}\] = 0.34 volt

Answer 1

Given:

\[\ce{Zn_{(s)} + Cu{^{2+}} (0.20 M) -> Zn{^{2+}} (0.50 M) + Cu_{(s)}}\]

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

\[\ce{E^{\circ}_{({Cu^{2+}/{Cu}})}}\] = 0.34 V

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

= 0.34 − (−0.76)

= 1.10 V

Use the Nernst equation:

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

Where:

n = 2 electrons exchanged

Zn²⁺ = 0.50 M

Cu²⁺ = 0.20 M

\[\ce{E_{cell} = 1.10 - \frac{0.0591}{2} log \frac{0.50}{0.20}}\]

= 1.10 − 0.02955 log (2.5)

= 1.10 − 0.02955 × 0.398    ...(log (2.5) ≈ 0.398)

= 1.10 − 0.01176

= 1.088 V