Question

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

\[\ce{Fe^2+_{( aq)} + Ag+_{( aq)} -> Fe^3+_{( aq)} + Ag_{(s)}}\]

Calculate the Δ_rG^θ and equilibrium constant of the reaction.

Answer 1

A cell can be represented as follows –

\[\ce{Fe^{2+}_{( aq)} | Fe^{3+}_{( aq)} || Ag^+_{( aq)} | Ag_{(s)}}\]

The cell reaction is as follows:

\[\ce{Fe^{2+}_{( aq)} + Ag^+_{( aq)} -> Fe^{3+}_{( aq)} + Ag_{(s)}}\]

So, n = 1

\[\ce{E^Θ_{{cell}} = E^Θ_{Ag^+/Ag} - E^Θ_{{Fe^{3+}/Fe^{2+}}}}\]

= +0.80 − (+0.77)

= +0.03 V

Δ_rG^Θ = `-"nFE"_"cell"^Θ`

= −1 × 96500 × 0.03

= −2895 CV mol⁻¹

= −2895 J mol⁻¹

= −2.895 kJ mol⁻¹

∵ Δ_rG^Θ = − 2.303 RT log₁₀ K

∴ −2895 = −2.303 × 8.314 × 298 × log₁₀ K

or, log₁₀ K = `2895/(2.303 xx 8.314 xx 298)` = 0.5074

∴ K = antilog (0.5074) = 3.22