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Q2.12(ii):
How much charge is required for the following reduction: 1 mol of \(Cu^{2+}\) to Cu?

Solution :

Step 1: Analyzing the Half-Reaction

To determine the charge required for the reduction of $Cu^{2+}$ to $Cu$, we must first establish the stoichiometry of the electron transfer. The reduction half-reaction is expressed as:

$Cu^{2+} (aq) + 2e^- \rightarrow Cu (s)$

[Per the principle of charge conservation in redox reactions, the change in oxidation state must be balanced by the gain of electrons.]

Step 2: Identifying Stoichiometric Requirements

From the balanced equation above, it is evident that 1 mole of $Cu^{2+}$ ions requires 2 moles of electrons ($e^-$) to be reduced to 1 mole of metallic copper ($Cu$).

  • Number of moles of electrons ($n$) = $2 \text{ mol}$

Step 3: Calculating Total Charge (Faraday's Law)

The total charge ($Q$) is related to the number of moles of electrons ($n$) and the Faraday constant ($F$). The Faraday constant represents the magnitude of electric charge per mole of electrons, approximately $96487 \text{ C mol}^{-1}$.

Formula: $Q = n \times F$

Substituting the given values:

$Q = 2 \text{ mol} \times 96487 \text{ C mol}^{-1}$

[By Faraday’s First Law of Electrolysis, the quantity of electricity required for reduction is proportional to the moles of electrons transferred.]

Step 4: Calculation and Unit Analysis

$Q = 192974 \text{ C}$

Cu²⁺ Cu + 2e⁻ Transfer Charge = 192,974 Coulombs

For standard academic reporting, rounding to 193,000 C is often acceptable, but the precise calculation yields 192,974 C.

Final Solution: The charge required for the reduction of 1 mol of $Cu^{2+}$ to $Cu$ is 192,974 C.


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