Question

Calculate the standard cell potentials of galvanic cells in which the following reactions take place:
(i) $2Cr_{(S)}+3Cd^{2+}_{(aq)} \rightarrow 2Cr^{3+}_{(aq)}+3Cd$
(ii) $Fe^{2+}_{(aq)}+Ag^+_{(aq)} \rightarrow Fe^{3+}_{(Aq)}+Ag(s)$
Calculate the $\triangle_rG^\ominus$ and equilibrium constant of the reactions.

Answer 1

(i) $E^{\ominus}_{Cr^{3+} / Cr} = 0.74 V$

$E^\ominus_{Cd^{2+}/ Cd} = -0.40 V$

The galvanic cell of the given reaction is depicted as:

$Cr_{(s)}|Cr^{3+}_{(aq)}||Cd^{2+}_{(aq)}|Cd_{(s)}$

Now, the standard cell potential is

$E^\ominus_{cell}$= $E^{\ominus}_{R}$-$E^\ominus_L$

=-0.40-(-0.74)

=+0.34 V

$\triangle_tG^\ominus = -nFE^\ominus_{cell}$

in the given equation,

n=6

F= 96487 C mol-1

$E^\ominus_{cell}$ = +0.34 V

Then, $\triangle_tG^\ominus$ = -6 * 96487 C mol-1 $\times$ 0.34 V

= - 196833.48 CV mol-1

= - 196833.48 j mol-1

= - 196.83 kj mol-1

Again

$\triangle_tG^\ominus$ = -RT in K

$\Rightarrow \triangle_tG^\ominus$ = - 2.303 R$$\Tau in K

$\Rightarrow$ log K = - $\frac{\triangle_tG^\ominus}{2.303RT}$

= $\frac{-196.83 \times 10^3}{2.303 \times 8.314 \times 298}$

= 34.496

∴ K = antilog (34.496) = 3.13 × 1034

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(ii) $E^\ominus$ Fe3+/Fe2+ = 0.77 V

$E^\ominus$ Ag+/Ag = -0.80 V

The galvanic cell of the given reaction is depicted as:

Fe2+ (aq) | Fe3+(aq)||Ag+(aq)|Ag(s)

Now, the standard cell potential is

$E^\ominus_{cell}$ = $E^{\ominus}_{R}$- $E^\ominus_L$

= 0.80-0.77

=0.03 V

Here, n = 1.

Then, $\triangle_tG^\ominus = -nFE^\ominus_{cell}$

= - 1 × 96487 C mol-1 × 0.03 V

= - 2894.61 J mol-1

= - 2.89 kJ mol-1

Again,

$\triangle_tG^\ominus$ = - 2.303 RT in K

$\Rightarrow$ log K = - $\frac{\triangle_tG^\ominus}{2.303RT}$

= $\frac{-2894.61}{2.303\times 8.314 \times 298}$

= 0.5073

∴ K = antilog (0.5073)

= 3.2 (approximately)