Question

A current of 10 A flows for 19.3 hours through an electrolytic cell containing a molten salt of metal X. This results in the decomposition of 2.4 mole of metal X at the cathode. The oxidation state of X in the molten salt is

• A. 1+
• B. 2+
• C. 3+
• D. 4+

Answer 1

Answer: (C)

3+

Answer 2

The deposition of a metal at the cathode during electrolysis follows Faraday's laws of electrolysis. According to Faraday's first law, the mass of a substance deposited is proportional to the charge passed. According to Faraday's second law, when the same quantity of electricity is passed through different electrolytes, the masses of the substances deposited are proportional to their equivalent weights.

The reduction reaction at the cathode is $X^{n+} + n e^- \rightarrow X$, where $n$ is the oxidation state of metal X in the molten salt. This means that $n$ moles of electrons are required to deposit 1 mole of metal X.

The total charge $q$ passed through the electrolytic cell is given by the product of the current $I$ and the time $t$. Given $I = 10$ A and $t = 19.3$ hours. First, convert the time to seconds: $t = 19.3 \text{ hours} \times 60 \text{ minutes/hour} \times 60 \text{ seconds/minute} = 19.3 \times 3600 \text{ s} = 69480 \text{ s}$.

Now, calculate the total charge passed: $q = I \times t = 10 \text{ A} \times 69480 \text{ s} = 694800 \text{ C}$.

The number of moles of electrons passed is given by $q/F$, where $F$ is Faraday's constant ($F \approx 96500$ C/mol). Number of moles of electrons $=\frac{q}{F} = \frac{694800 \text{ C}}{96500 \text{ C/mol}}$.

From the reduction reaction $X^{n+} + n e^- \rightarrow X$, $n$ moles of electrons deposit 1 mole of X. Therefore, 1 mole of electrons deposits $1/n$ moles of X. The total number of moles of X deposited, $N$, is equal to the total number of moles of electrons passed divided by $n$. $N = \frac{\text{Number of moles of electrons}}{n} = \frac{q/F}{n} = \frac{q}{nF}$.

We are given that $N = 2.4$ moles of metal X are deposited. So, $2.4 = \frac{694800}{n \times 96500}$.

Now, we can solve for $n$, the oxidation state of X: $n = \frac{694800}{2.4 \times 96500}$. $n = \frac{694800}{231600}$. $n = \frac{6948}{2316}$.

Performing the division: $6948 \div 2316 = 3$.

So, the oxidation state of X in the molten salt is +3.