JEE Chemistry 2023 Question with Solution

Given: $E^{0}_{X^{2+} \mid X} = -2.36 \, \text{V}$, $E^{0}_{Y^{2+} \mid Y} = +0.36 \, \text{V}$, $[X^{2+}] = 0.001 \, \text{M}$, $[Y^{2+}] = 0.01 \, \text{M}$ and $\frac{2.303RT}{F} = 0.06 \, \text{V}$.

Find: The electrode potential in the form $n \times 10^{-2} \, \text{V}$.

The cell reaction used in the solution is

$$X + Y^{2+} \rightarrow Y + X^{2+}$$

So,

$$E^{0}_{\text{cell}} = 0.36 - (-2.36) = 2.72 \, \text{V}$$

Using the Nernst equation as shown,

$$E_{\text{cell}} = 2.72 - \frac{0.06}{2} \log \left(\frac{0.01}{0.001}\right)$$

Since

$$\log\left(\frac{0.01}{0.001}\right) = \log(10) = 1$$

we get

$$E_{\text{cell}} = 2.72 + 0.03 = 2.75 \, \text{V}$$

Therefore,

$$2.75 \, \text{V} = 275 \times 10^{-2} \, \text{V}$$

So, the required nearest integer is 275.

Given: The final cell potential obtained from the Nernst equation is $2.75 \, \text{V}$.

Find: The integer multiplying $10^{-2} \, \text{V}$.

Rewrite the voltage as

$$2.75 \, \text{V} = 2.75 \times 100 \times 10^{-2} \, \text{V}$$ $$= 275 \times 10^{-2} \, \text{V}$$

Hence the nearest integer asked in the question is 275.