Question

A current of $2.0\text{ A}$ is passed for 5 hours through an electrolytic cell containing an aqueous solution of a metal salt, depositing $12.0\text{ g}$ of the metal at the cathode. If the atomic mass of the metal is $193\text{ g mol}^{-1}$, find the oxidation state of the metal ion in the solution. (Take Faraday's constant $F = 96500\text{ C mol}^{-1}$).

• A. \( +1 \)
• B. \( +2 \)
• C. \( +3 \)
• D. \( +6 \)

Hint:

A fast alternative method is: \[ \text{Moles of metal deposited} = \frac{12}{193} \] \[ \text{Moles of electrons passed} = \frac{36000}{96500} \] Then, \[ n = \frac{\text{moles of electrons}}{\text{moles of metal}} \] which again gives: \[ n \approx 6 \]

Answer 1

The Correct Option is D

Concept: According to Faraday's First Law of Electrolysis: \[ m = \frac{MIt}{nF} \] where:
• \(m\) = mass deposited
• \(M\) = molar mass
• \(I\) = current
• \(t\) = time
• \(F\) = Faraday constant
• \(n\) = oxidation state (valency)

Step 1: Calculate total charge passed.
Given: \[ I = 2.0\text{ A}, \qquad t = 5\text{ h} \] Convert time into seconds: \[ t = 5 \times 60 \times 60 = 18000\text{ s} \] Therefore, \[ Q = It = 2.0 \times 18000 = 36000\text{ C} \]

Step 2: Use Faraday's formula to find \(n\).
Given: \[ m = 12.0\text{ g}, \qquad M = 193\text{ g mol}^{-1} \] Using: \[ m = \frac{MIt}{nF} \] Rearranging: \[ n = \frac{MIt}{mF} \] Substitute the values: \[ n = \frac{193 \times 2.0 \times 18000}{12.0 \times 96500} \]

Step 3: Simplify the expression.
\[ n = \frac{193 \times 36000}{12 \times 96500} \] \[ n = \frac{6,948,000}{1,158,000} \] \[ n = 6 \] Hence, the oxidation state of the metal ion is: \[ \boxed{+6} \]