Question

For the following cell, calculate the emf:

\[\ce{Al | Al^{3+} (0.01 M) || Fe^{2+} (0.02 M) | Fe}\]

Given: \[\ce{E^{\circ}_{{Al^{3+}/{Al}}}}\] = −1.66 V, \[\ce{E^{\circ}_{{Fe^{2+}/{Fe}}}}\] = −0.44 V

Answer 1

Given:

\[\ce{Al | Al^{3+} (0.01 M) || Fe^{2+} (0.02 M) | Fe}\]

\[\ce{E^{\circ}_{{Al^{3+}/{Al}}}}\] = −1.66 V

\[\ce{E^{\circ}_{{Fe^{2+}/{Fe}}}}\] = −0.44 V

To Find:

Emf of cell = ?

Calculations:

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

= −0.44 − (−1.66)

= 1.22 V

Using the Nernst equation:

The balanced redox reaction is:

\[\ce{2Al + 3Fe^{2+} −> 2Al^{3+} + 3Fe}\]

n = 6 electrons exchanged (LCM of 3 for Al³⁺ and 2 for Fe²⁺)

Al³⁺ = 0.01 M, Fe²⁺ = 0.02 M

\[\ce{E_{cell} = E{^{\circ}_{cell}} - \frac{0.0591}{n} log \frac{[Al^{3+}]^2}{[Fe^{2+}]^3}}\]

= \[\ce{1.22 - \frac{0.0591}{6} log \frac{(0.01)^2}{(0.02)^3}}\]

= \[\ce{1.22 - 0.00985 log \frac{10^{-4}}{8 \times 10^{-6}}}\]

= 1.22 − 0.00985 log (12.5)

= 1.22 − 0.00985 × 1.0969    ...(log 12.5 ≈ 1.0969)

= 1.22 − 0.0108

= 1.209 V