Question

Calculate the standard cell potential of a galvanic cell in which the following reaction takes place:

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

Calculate the Δ_rG° and equilibrium constant of the reaction.

Calculate Δ_rG° and log K_C of the reaction:

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

Given: \[\ce{E{^{\circ}_{{Cr^{3+}/Cr}}} = -0.74 V E{^{\circ}_{{Cd^{2+}/Cd}}} = -0.40 V}\]

[R = 8.314 J K⁻¹ mol⁻¹, F = 96500 C mol⁻¹]

Answer 1

A cell can be represented as follows:

\[\ce{Cr_{(s)} | Cr{^{3+}_{(aq)}} || Cd{^{2+}_{(aq)}} | Cd_{(s)}}\]

\[\ce{E{^{\circ}_{{cell}}} = E{^{\circ}_{R}} - E{^{\circ}_{L}}}\]

= \[\ce{E{^{\circ}_{{Cd^{2+}/Cd}}} - E{^{\circ}_{{Cr^{3+}/Cr}}}}\]

= − 0.40 − (− 0.74)

= + 0.34 V

∴ \[\ce{\Delta_{r}G^{\circ} = -nFE{^{\circ}_{cell}}}\]

= −6 × 96500 × 0.34    ...(F = 96500 C mol⁻¹)

= −196860 CV mol⁻¹

= −196860 J mol⁻¹

= −196.86 kJ mol⁻¹

∵ Δ_rG° = −2.303 RT log₁₀ K_C

∴ −196860 = −2.303 × 8.314 × 298 × log₁₀ K_C

or, log₁₀ K_C = `196860/(2.303 xx 8.314 xx 298)`

= 34.5014

∴ K_C = antilog₁₀ (34.5014)

= 3.172 × 10³⁴