Question

When the same quantity of electricity is passed through the aqueous solutions of the given electrolytes for the same amount of time, which metal will be deposited in maximum amount on the cathode?

• A. ZnSO$_4$
• B. FeCl$_3$
• C. AgNO$_3$
• D. NiCl$_2$

Hint:

The metal with the highest equivalent weight gets deposited in the maximum amount when the same charge is passed through different electrolytes.

Answer 1

The Correct Option is C

Step 1: Understanding Faraday’s Laws of Electrolysis 
According to Faraday’s first law of electrolysis, the mass of metal deposited on the cathode is directly proportional to the charge passed: \[ m = \frac{ZQ}{F} \] where: \( m \) is the mass of the deposited metal, 
\( Z \) is the electrochemical equivalent of the metal, 
\( Q \) is the charge passed (which is the same for all solutions), 
\( F \) is Faraday’s constant \( 96500 \) C/mol.
Step 2: Electrochemical Equivalent and Equivalent Weight 
The electrochemical equivalent (\( Z \)) is given by: \[ Z = \frac{{Atomic weight}}{{Valency} \times F} \] The more the electrochemical equivalent (\( Z \)), the greater the mass of metal deposited. 
The equivalent weight is given by:
\[ {Equivalent weight} = \frac{{Atomic weight}}{{Valency}} \] 
Step 3: Calculating Equivalent Weights for Given Options 
Zn from ZnSO$_4$ - Atomic weight of Zn = 65 - Valency = 2 - Equivalent weight = \( \frac{65}{2} = 32.5 \) 
Fe from FeCl$_3$ - Atomic weight of Fe = 56 - Valency = 3 - Equivalent weight = \( \frac{56}{3} \approx 18.67 \) 
Ag from AgNO$_3$ - Atomic weight of Ag = 108 - Valency = 1 - Equivalent weight = \( \frac{108}{1} = 108 \) 
Ni from NiCl$_2$ - Atomic weight of Ni = 58.7 - Valency = 2 - Equivalent weight = \( \frac{58.7}{2} = 29.35 \) 
Step 4: Identifying the Metal Deposited in Maximum Amount 
- Since the mass deposited is directly proportional to equivalent weight, the metal with the highest equivalent weight will be deposited in the maximum amount. - Silver (Ag) has the highest equivalent weight (108 g/mol), so Ag will be deposited in the greatest amount. Thus, the correct answer is (C) AgNO$_3$.