Question

Calculate the emf and ΔG for the given cell at 25°C:

Cr_(s) | Cr³⁺ (0.1 M) || Fe²⁺ (0.01 M) || Fe_(s)

Given: \[\ce{E^{\circ}_{Cr^{3+}/Cr}}\] = −0.74 V, \[\ce{E^{\circ}_{Fe^{2+}/Fe}}\] = −0.44 V

(1 F = 96500 C mol⁻¹, R = 8.314 JK⁻¹ mol⁻¹)

Answer 1

The given cell is:

Cr_(s) | Cr³⁺ (0.1 M) || Fe²⁺ (0.01 M) || Fe_(s)

Anode (oxidation): \[\ce{Cr -> Cr^{3+} + 3e-}\] = −0.74 V

Cathode (reduction): \[\ce{Fe^{2+} + 2e- -> Fe}\] = −0.44 V

We know that,

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

= (−0.44) − (−0.74)

= +0.30 V

To balance electrons multiply Fe half-reaction by 3, Cr half-reaction by 2.

\[\ce{2Cr_{(s)} + 3Fe^{2+}_{ (aq)} -> 2Cr^{3+}_{ (aq)} + Fe_{(s)}}\]

Total electrons transferred (n) = 6

Using the Nernst equation:

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

= \[\ce{0.30 - \frac{0.0591}{6} log \frac{(0.1)^2}{(0.01)^3}}\]

= \[\ce{0.30 - \frac{0.0591}{6} log \frac{10^{-2}}{10^{-6}}}\]

= \[\ce{0.30 - \frac{0.0591}{6} log (10^4)}\]

= \[\ce{0.30 - \frac{0.0591}{6} \times 4}\]

= 0.30 − 0.0394

E_cell = 0.2606 V

ΔG = −nFE_cell​

= −6 × 96500 × 0.2606

= −150843.9 J/mol

= −150.84 kJ/mol