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^Θ and equilibrium constant of the reaction.

Answer 1

A cell can be represented as follows –

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

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

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

= − 0.40 − (− 0.74)

= + 0.34 V

∴ Δ_rG^Θ = `-"nFE"_"cell"^Θ`

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

= −196860 CV mol⁻¹

= −196860 J mol⁻¹

= −196.86 kJ mol⁻¹

∵ Δ_rG^Θ = −2.303 RT log₁₀ K

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

or, log₁₀ K = `196860/(2.303 xx 8.314 xx 298)` = 34.5014

∴ K = antilog₁₀ (34.5014) = 3.172 × 10³⁴