Question

A current of 1.5 A is passed for 2 hours through an aqueous solution of PdX\(_n\), where X is a monovalent anion. During the electrolysis process 2.977 g of Palladium metal gets deposited at the cathode. Calculate the charge on Pd ions. (Atomic mass of Pd = 106.4 g/mol).

• A. \( n = 4 \)
• B. \( n = 6 \)
• C. \( n = 3 \)
• D. \( n = 2 \)

Hint:

In electrolysis problems, use the formula \( m = \frac{M \cdot I \cdot t}{n \cdot F} \) to calculate the amount of substance deposited at the cathode. Remember that \( n \) corresponds to the number of electrons involved in the reduction half-reaction.

Answer 1

The Correct Option is A

Step 1: Use the formula for the amount of substance deposited in electrolysis.
The amount of substance deposited at the cathode is given by the formula:
\[ m = \frac{M \cdot I \cdot t}{n \cdot F} \]
where:
- \( m \) is the mass of the substance deposited,
- \( M \) is the molar mass of the substance (Palladium, \( M = 106.4 \, \text{g/mol} \)),
- \( I \) is the current (1.5 A),
- \( t \) is the time (2 hours = 7200 s),
- \( n \) is the number of electrons involved in the reduction (which we are solving for),
- \( F \) is the Faraday constant (\( F = 96500 \, \text{C/mol} \)).

Step 2: Rearrange the formula to solve for \( n \).
Rearranging the formula to solve for \( n \): \[ n = \frac{M \cdot I \cdot t}{m \cdot F} \]

Step 3: Substitute the known values.
Substitute the known values into the equation:
\[ n = \frac{106.4 \times 1.5 \times 7200}{2.977 \times 96500} \]
\[ n = \frac{1064.4 \times 7200}{288,000.5} \]
\[ n = \frac{7,659,168}{288,000.5} \approx 4 \]

Step 4: Conclusion.
Thus, the number of electrons involved in the reduction of Pd\(^{2+}\) to Pd is \( n = 4 \), and the correct answer is option (A).