Question

\(Al_2O_3\) can be electrolyzed with an inert anode or a carbon anode.
Overall reactions are as follows:
Reaction I: For inert anode,
\[ \frac{2}{3}Al_2O_3(s) \to \frac{4}{3}Al(l) + O_2(g); \quad \Delta G^\circ \text{ (in Joules)} = 1124800 - 218T \]
Reaction II: For carbon anode,
\[ \frac{2}{3}Al_2O_3(s) + C(s) \to \frac{4}{3}Al(l) + CO_2(g); \quad \Delta G^\circ \text{ (in Joules)} = 730700 - 218T \]
T denotes temperature in Kelvin.
Given: Faraday constant = 96500 Coulomb.
If the inert anode is replaced by a carbon anode during electrolysis of \(Al_2O_3\) at temperature 1300 K, the decrease in the magnitude of decomposition potential between the two reactions (rounded off to two decimal places) is _________________ Volts.

Hint:

Find n=4 electrons from the 4/3 mol of Al produced, convert each ΔG° into E = -ΔG°/(nF) at T=1300 K, then take the difference in magnitude.

Answer 1

Correct Answer: 1

Step 1: Understanding the Concept:
Using a carbon anode instead of an inert one changes the overall cell reaction, because the anode itself now reacts with the oxygen released, forming $CO_2$ instead of $O_2$. This lowers the Gibbs free energy needed to drive the electrolysis, and hence lowers the decomposition potential.

Step 2: Key Formula or Approach:
Rather than working out $E_1$ and $E_2$ separately from scratch, we can find the drop in Gibbs energy directly and divide once by $nF$, since $n$ and $F$ are the same in both reactions.
\[ \Delta(\Delta G^\circ) = \Delta G_1^\circ - \Delta G_2^\circ \]
\[ \Delta E = \frac{\Delta(\Delta G^\circ)}{nF} \]

Step 3: Detailed Explanation:
At $T=1300$ K,
\[ \Delta G_1^\circ = 1124800-218(1300)=841400 \text{ J} \]
\[ \Delta G_2^\circ = 730700-218(1300)=447300 \text{ J} \]
The drop between the two is
\[ \Delta(\Delta G^\circ) = 841400-447300 = 394100 \text{ J} \]
Each reaction transfers $n=4$ electrons per mole of reaction as written, since $\frac{4}{3}$ mol of Al needs $4$ mol of electrons. So
\[ \Delta E = \frac{394100}{4\times96500} = \frac{394100}{386000} = 1.0210 \text{ V} \]

Step 4: Final Answer:
Switching to the carbon anode drops the decomposition potential by about $1.02$ V.
\[ \boxed{\Delta E \approx 1.02 \text{ V}} \]