Question

Calculate \(\Delta_r G^{\circ}\) and \(\log K_C\) of the reaction.

\(\mathrm{2Cr(s) + 3Cd^{2+}(aq) \to 2Cr^{3+}(aq) + 3Cd(s)}\)

Given \(E^{\circ}_{\mathrm{Cr^{3+}/Cr}} = -0.74\) V, \(E^{\circ}_{\mathrm{Cd^{2+}/Cd}} = -0.40\) V

[R = 8.314 J \(\mathrm{K^{-1}\,mol^{-1}}\), F = 96500 C \(\mathrm{mol^{-1}}\)]

Hint:

\(E^{\circ}_{\mathrm{cell}} = -0.40 - (-0.74) = 0.34\) V and \(n = 6\).
\(\Delta_r G^{\circ} = -nFE^{\circ}\) and \(\Delta_r G^{\circ} = -2.303RT\log K_C\).

Answer 1

Step 1: Concept:
The standard emf of a cell tells us how far a reaction goes. A positive emf gives a negative $\Delta_r G^{\circ}$ and a large equilibrium constant.

Step 2: Identify the cathode and the anode:
Cadmium has the less negative electrode potential (-0.40 V), so $Cd^{2+}$ is reduced and is the cathode. Chromium (-0.74 V) is oxidised and is the anode.
Cr goes from 0 to +3, losing 3 electrons, and for 2 Cr atoms the total is 6 electrons. So $n = 6$.

Step 3: Emf of the cell:
$E^{\circ}_{cell} = (-0.40) - (-0.74) = 0.34$ V.

Step 4: Find the standard Gibbs energy:
Use $\Delta_r G^{\circ} = -nFE^{\circ}_{cell}$.
\[ \Delta_r G^{\circ} = -(6)(96500\ \text{C mol}^{-1})(0.34\ \text{V}) = -1.9686 \times 10^{5}\ \text{J mol}^{-1} \]

Step 5: Find log Kc from the Gibbs energy:
At equilibrium $\Delta_r G^{\circ} = -RT \ln K_C$, and $\ln K = 2.303 \log K$.
\[ \log K_C = \frac{1.9686\times10^{5}}{2.303 \times 8.314 \times 298} \]
The denominator is $2.303 \times 8.314 = 19.147$, and $19.147 \times 298 = 5705.8$. Dividing gives $34.5$.

Final Answer:
$\Delta_r G^{\circ} = -196.86$ kJ per mol and $\log K_C = 34.5$.\[ \boxed{\Delta_r G^{\circ} = -1.97\times10^{5}\ \text{J mol}^{-1},\ \log K_C = 34.5} \]