Q11 / 180 · COMEDK 2026 (9 May Shift 1) · Chemistry
$3.482 \times 10^{-1}$g of Fe gets deposited when an aqueous solution of Ferric sulphate is electrolysed for 20 minutes using a current of "x" amperes. Find "x". (Atomic mass of Fe = 56amu).
- Correct answerA$1.5\text{A}$
- B$2.4\text{A}$
- C$0.98\text{A}$
- D$2.8\text{A}$
Solution
The chemical formula of ferric sulphate is $\text{Fe}_2(\text{SO}_4)_3$. The reduction reaction at the cathode is:
$\text{Fe}^{3+} + 3e^{-} \rightarrow \text{Fe}$
The number of electrons involved in the reduction of one $\text{Fe}^{3+}$ ion is $3$. Thus, the equivalent mass of iron is $E = \dfrac{56}{3}$.
According to Faraday's first law of electrolysis:
$W = \dfrac{E \cdot I \cdot t}{96500}$
Given:
Mass deposited, $W = 3.482 \times 10^{-1} \text{ g} = 0.3482 \text{ g}$
Time, $t = 20 \text{ minutes} = 20 \times 60 = 1200 \text{ s}$
Current, $I = x$
Substituting the values into the formula:
$0.3482 = \dfrac{56 \cdot x \cdot 1200}{3 \cdot 96500}$
$0.3482 = \dfrac{67200 \cdot x}{289500}$
$x = \dfrac{0.3482 \cdot 289500}{67200}$
$x = \dfrac{100803.9}{67200} = 1.5 \text{ A}$
Answer: $1.5\text{A}$
$\text{Fe}^{3+} + 3e^{-} \rightarrow \text{Fe}$
The number of electrons involved in the reduction of one $\text{Fe}^{3+}$ ion is $3$. Thus, the equivalent mass of iron is $E = \dfrac{56}{3}$.
According to Faraday's first law of electrolysis:
$W = \dfrac{E \cdot I \cdot t}{96500}$
Given:
Mass deposited, $W = 3.482 \times 10^{-1} \text{ g} = 0.3482 \text{ g}$
Time, $t = 20 \text{ minutes} = 20 \times 60 = 1200 \text{ s}$
Current, $I = x$
Substituting the values into the formula:
$0.3482 = \dfrac{56 \cdot x \cdot 1200}{3 \cdot 96500}$
$0.3482 = \dfrac{67200 \cdot x}{289500}$
$x = \dfrac{0.3482 \cdot 289500}{67200}$
$x = \dfrac{100803.9}{67200} = 1.5 \text{ A}$
Answer: $1.5\text{A}$
Hence, the correct answer is option: 'A'.