Question

For the given reactions

\[\ce{Sn^{+2} + 2e- -> Sn}\]

\[\ce{Sn^{+4} + 2e- -> Sn}\]

The electrode potentials are \[\ce{E^{\circ}_{{Sn^{+2}/{Sn}}}}\] = −0.14 V and \[\ce{E^{\circ}_{{Sn^{+4}/{Sn}}}}\] = 0.010 V. The magnitude of standard electrode potential for \[\ce{Sn^{+4}/Sn^{+2}}\] i.e., \[\ce{E^{\circ}_{{Sn^{+4}/{Sn^{+2}}}}}\] is ______ × 10⁻² V (Nearest integer).

Answer 1

The electrode potentials are \[\ce{E^{\circ}_{{Sn^{+2}/{Sn}}}}\] = −0.14 V and \[\ce{E^{\circ}_{{Sn^{+4}/{Sn}}}}\] = 0.010 V. The magnitude of standard electrode potential for \[\ce{Sn^{+4}/Sn^{+2}}\] i.e., \[\ce{E^{\circ}_{{Sn^{+4}/{Sn^{+2}}}}}\] is 15 × 10⁻² V.

Explanation:

We are given:

\[\ce{E^{\circ}_{{Sn^{2+}/{Sn}}}}\] = −0.14 V

\[\ce{E^{\circ}_{{Sn^{4+}/{Sn}}}}\] = +0.010 V

We can combine two half-reactions:

\[\ce{Sn^{4+} + 4e− −> Sn}\], E° = +0.010 V    ...(i)

\[\ce{Sn{^{2+}} + 2e− −> Sn}\], E° = −0.14 V    ...(ii)

Now we can get the equation for:

\[\ce{Sn{^{4+}} + 2e− −>Sn^{2+}}\]

This is obtained by subtracting reaction (ii) from (i) using the formula based on standard Gibbs free energy.

We know that,

ΔG° = −nFE°

So,

\[\ce{\Delta G^{\circ}_{{Sn^{4+}/{Sn}}} = \Delta G^{\circ}_{{Sn^{4+}/{Sn^{2+}}}} + \Delta G^{\circ}_{{Sn^{2+}/{Sn}}}}\]

Convert to potentials:

\[\ce{\Delta G^{\circ}_{{Sn^{4+}/{Sn}}} = \frac{n_1E{^{\circ}_{1}} + n_2E{^{\circ}_{2}}}{n_1 + n_2}}\]

\[\ce{E^{\circ}_{{Sn^{4+}/{Sn^{2+}}}} = E^{\circ}_{{Sn^{4+}/{Sn}}} - E^{\circ}_{{Sn^{2+}/{Sn}}}}\]

\[\ce{E^{\circ}_{{Sn^{4+}/{Sn^{2+}}}}}\] = 0.010 − (−0.14)

= 0.150 V

= 15 × 10⁻² V