Question

How much charge in faraday is required for the reduction of 1 mol of Ag$^+$ to Ag?

Hint:

The charge required in Faraday can be directly found from the number of electrons involved: \[ \boxed{\text{Charge (F)} = \text{Number of moles of electrons}} \] For: \[ M^{n+}+ne^- \rightarrow M \] the required charge is: \[ \boxed{nF} \] For Ag$^+$: \[ n=1 \Rightarrow \boxed{1F} \]

Answer 1

Step 1: Write the electrode reaction.
Silver ions are reduced at the cathode: \[ Ag^+ + e^- \rightarrow Ag \] The stoichiometry shows one electron is needed per $Ag^+$ ion.
Step 2: Scale to 1 mole of $Ag^+$.
1 mole of $Ag^+$ requires 1 mole of electrons.
Step 3: Convert to Faraday.
By definition, 1 Faraday is the charge carried by 1 mole of electrons ($= 96485$ C). Therefore the charge required for reduction of 1 mol $Ag^+$ is exactly 1 Faraday. \[ \boxed{1\text{ Faraday}} \]