Question

\[\ce{Cu_{(s)} + Sn{^{+2}} (0.001 M) -> Cu^{+2} (0.01 M) + Sn_{(s)}}\]

The Gibbs free energy change for the above reaction at 298 K is x × 10⁻¹ KJ mol⁻¹. The value of x is ______ (Nearest integer).

(Given \[\ce{E^{\circ}_{{Cu^{2+}/{Cu}}} = 0.34 V}\], \[\ce{E^{\circ}_{{Sn^{2+}/{Sn}}} = -0.14 V}\], F = 96500 C mol⁻¹)

Answer 1

The value of x is 984.

Explanation:

The given cell reaction is

\[\ce{Cu_{(s)} + Sn{^{+2}} (0.001 M) -> Cu^{+2} (0.01 M) + Sn_{(s)}}\]

The half cell reactions are:

Anode (oxidation):
\[\ce{Cu -> Cu^{2+} + 2e-}\], E° = +0.34 V

Cathode (reduction):
\[\ce{Sn^{2+} + 2e− -> Sn}\], E° = −0.14 V

So,

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

= −0.14 − 0.34

= −0.48 V

Reaction quotient Q = `\frac{[Cu^{2+}]}{[Sn^{2+}]}`

= `0.01/0.001`

= 10

Use the Nernst equation:

\[\ce{E = E{^{\circ}_{cell}} - \frac{0.0591}{n} log Q}\]

= `-0.48 - 0.0592/2 * log(10)`

= −0.48 − 0.0296

= −0.5096 V

ΔG = −nFE

= −2 × 96500 × (−0.5096)

= +98352.8 J mol⁻¹

= 98.3528 kJ mol⁻¹

ΔG = x × 10⁻¹

x = `98.3528/0.1`

x = 984