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° 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

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

= +0.80 − (+0.77)

= +0.03 V

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

= −1 × 96500 × 0.03

= −2895 CV mol⁻¹

= −2895 J mol⁻¹

= −2.895 kJ mol⁻¹

∵ Δ_rG° = − 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)

K = 3.22