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

Concept: According to Faraday's laws of electrolysis, the amount of substance deposited or reduced at an electrode depends on the quantity of electricity passed through the electrolyte. The charge required is calculated using the stoichiometry of electrons involved in the electrode reaction.

Step 1: Write the reduction reaction. Silver ions undergo reduction at the cathode: \[ Ag^+ + e^- \rightarrow Ag \] Here, one mole of Ag$^+$ ions accepts one electron to form one mole of silver metal.

Step 2: Calculate the number of electrons required. From the reaction: \[ 1 \text{ mol of } Ag^+ \text{ requires } 1 \text{ mol of } e^- \] Therefore: \[ \text{Moles of electrons required}=1 \]

Step 3: Convert electrons into Faraday. One Faraday is the charge carried by one mole of electrons: \[ 1F = 1 \text{ mol of } e^- \] Hence: \[ \text{Charge required}=1F \]

Final Answer: For the reduction of 1 mol of Ag$^+$ to Ag: \[ \boxed{ Ag^+ + e^- \rightarrow Ag } \] 1 mole of electrons is required, therefore the charge required is: \[ \boxed{1 \text{ Faraday}} \]