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

Solution :

Given Variables & Chemical Equation

The reduction reaction of copper(II) ions ($Cu^{2+}$) to metallic copper ($Cu$) is represented by the following half-reaction:

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

From the stoichiometric coefficients of the balanced half-reaction, it is observed that $1 \text{ mole}$ of $Cu^{2+}$ ions requires $2 \text{ moles}$ of electrons ($e^{-}$) to be reduced to $1 \text{ mole}$ of $Cu(s)$.

Step 1: Establishing the Relationship between Moles of Electrons and Charge

According to Faraday's First Law of Electrolysis, the total charge ($Q$) passed is related to the number of moles of electrons ($n$) by Faraday's constant ($F$).

$Q = n \times F$

  • $n = 2 \text{ mol}$ (Number of electrons transferred)
  • $F \approx 96,487 \text{ C mol}^{-1}$ (Commonly approximated as $96,500 \text{ C mol}^{-1}$ in academic problems)

[Per the definition of Faraday’s constant, $F = N_A \times e$, representing the charge of one mole of electrons.]

Step 2: Calculating the Total Charge

Using the values identified in Step 1:

$Q = 2 \text{ mol} \times 96,500 \text{ C mol}^{-1}$

$Q = 193,000 \text{ C}$

Visual Representation of Reduction

Cu²⁺ 2e⁻ Cu

Summary Table of Parameters

Parameter Value
Stoichiometric $e^{-}$ requirement $2 \text{ moles}$
Faraday's Constant ($F$) $96,500 \text{ C mol}^{-1}$
Total Charge ($Q$) $193,000 \text{ C}$

Final Solution: The charge required for the reduction of 1 mol of $Cu^{2+}$ to $Cu$ is $193,000 \text{ Coulombs (C)}$.


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