(b) (i) How much charge in faraday is required for the reduction of 1 mol of Ag⁺ to Ag?

Step 1 (Governing Reaction Equation): [1 Mark]

The reduction of silver ions ($Ag^+$) to metallic silver ($Ag$) involves the gain of one electron per ion. The balanced half-cell reaction is:

Step 2 (Step-by-Step Mathematical Working): [2 Marks]

• From the stoichiometry of the balanced equation, 1 mole of $Ag^+$ ions requires 1 mole of electrons ($e^-$) to be reduced to 1 mole of $Ag$.
• The charge carried by 1 mole of electrons is defined as 1 Faraday ($F$).
• Mathematically: $$Q = n \times F$$ Where $n$ is the number of moles of electrons transferred. $$n = 1 \text{ mol}$$ $$Q = 1 \times 1F = 1F$$

Step 3 (Final Concluded Answer): [1 Mark]

The total charge required for the reduction of 1 mole of $Ag^+$ to $Ag$ is 1 Faraday (1 F).

💡 EL10 Concept Insight:

Think of a Faraday as a "package deal" for electricity; since each silver ion only needs one single electron to become neutral, and one mole of electrons is exactly one Faraday, you simply need one Faraday of charge to finish the job.