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$_3$.

Answer 2

Step 1: Understanding Faraday’s Laws of Electrolysis
Faraday’s first law states that the mass of a substance deposited or liberated at an electrode during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte.
Faraday’s second law states that when the same quantity of electricity passes through different electrolytes, the amounts of substances deposited are proportional to their equivalent weights.

Step 2: Expression for Mass Deposited
The mass (m) of metal deposited at the cathode is given by:
m = (Q × M) / (n × F), where:
- Q = total charge passed
- M = molar mass of the metal
- n = number of electrons involved in reducing one metal ion
- F = Faraday constant (approximately 96500 C/mol)

Step 3: Comparing Different Metals
For the same quantity of electricity (Q), the mass deposited depends on M/n.
- Silver (Ag) in AgNO₃ has M ≈ 108 g/mol and n = 1
- Other metals like Cu (M ≈ 63.5 g/mol, n = 2), Zn (M ≈ 65.4 g/mol, n = 2), and Al (M ≈ 27 g/mol, n = 3) have lower M/n ratios.
Higher the M/n ratio, greater is the mass deposited for the same charge.

Step 4: Conclusion
Since Ag has a high molar mass and requires only one electron to reduce (n=1), the value of M/n is highest for silver among common metals.
Therefore, when the same quantity of electricity is passed through aqueous solutions of various metal salts for the same time, maximum amount of silver metal will be deposited from AgNO₃ solution.