Question

The standard electrode potential for Sn⁺⁴/Sn⁺² couple is +0.15 V and that for the Cr⁺³/Cr couple is −0.74 V. These two couples in their standard states are connected to make a cell. (1 Faraday = 96500 mol⁻¹)

What will be the value of standard Gibbs energy (ΔG°)?

Options

• −650.3 kJ

• −515.3 kJ

• −226.4 kJ

• −406.8 kJ

Answer 1

−515.3 kJ

Explanation:

Given: \[\ce{E^{\circ}_{Sn^{4+}/Sn^{2+}}}\] = +0.15 V

\[\ce{E^{\circ}_{Cr^{3+}/Cr}}\] = −0.74 V

F = 96500 C/mol

Cr/Cr³⁺ has a more negative E°, so it acts as an anode (oxidation).

Sn⁺⁴/Sn⁺² is the cathode (reduction).

We know that

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

= 0.15 − (−0.74)

= 0.15 + 0.74

= 0.89 V

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

\[\ce{Cr -> Cr^{3+} + 3e-}\]

To balance electrons, LCM of 2 and 3 = 6 electrons

n = 6

By using the formula, 

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

= −6 × 96500 × 0.89

= −515.3 kJ/mol