JEE Chemistry 2023 Question with Solution

Given: $E_{\text{cell}} = 0.712 \, \text{V}$, $E^\circ_{\mathrm{Fe^{3+}/Fe^{2+}}} = 0.771 \, \text{V}$, $E^\circ_{\mathrm{H^+/H_2}} = 0 \, \text{V}$, and $T = 298 \, \text{K}$.

Find: The ratio $\dfrac{[\mathrm{Fe^{2+}}]}{[\mathrm{Fe^{3+}}]}$.

The cell reaction used in the solution is

$$\frac{1}{2}H_2(g) + Fe^{3+}(aq) \rightarrow H^{+}(aq) + Fe^{2+}(aq)$$

For this reaction,

$$E^\circ_{\text{cell}} = E^\circ_{\mathrm{Fe^{3+}/Fe^{2+}}} - E^\circ_{\mathrm{H^+/H_2}} = 0.771 - 0 = 0.771 \, \text{V}$$

Using the Nernst equation as shown in the solution,

$$E = E^\circ - 0.059 \log \left(\frac{[Fe^{3+}]}{[Fe^{2+}]}\right)$$

Substituting the given value of cell potential,

$$0.712 = 0.771 - 0.059 \log \left(\frac{[Fe^{3+}]}{[Fe^{2+}]}\right)$$

Therefore,

$$\log \left(\frac{[Fe^{3+}]}{[Fe^{2+}]}\right) = \frac{0.059}{(0.771 - 0.712)} = 1$$

Hence,

$$\frac{[Fe^{3+}]}{[Fe^{2+}]} = 10$$

So,

$$\frac{[Fe^{2+}]}{[Fe^{3+}]} = 10$$

Therefore, the required nearest integer is $10$.

Given: $E_{\text{cell}} = 0.712 \, \text{V}$ and $E^\circ_{\text{cell}} = 0.771 \, \text{V}$.

Find: The concentration ratio asked in the question.

Write the Nernst equation:

$$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log Q$$

Here, from the extracted solution,

$$Q = \frac{[\mathrm{Fe^{3+}}]}{[\mathrm{Fe^{2+}}]}$$

and $n = 1$.

So,

$$0.712 = 0.771 - 0.0591 \log Q$$

Rearranging,

$$\log Q = \frac{0.771 - 0.712}{0.0591} \approx 1$$

Hence,

$$Q = 10$$

The extracted solution then reports the final required ratio as $10$. Therefore, the answer recorded from the solution is $10$.