JEE Chemistry 2026 Question with Solution

Given: $K_{sp} = 10^{-10}$ and $E^{\circ}_{M^{+}/M} = 0.79 \, \text{V}$.

Find: The standard reduction potential of the metal/metal insoluble salt electrode $E^{\circ}_{X^{-}/MX/M}$ in mV.

The solution states the relation

$$E^{\circ}_{X^{-}/MX/M} = E^{\circ}_{M^{+}/M} + \frac{0.059}{n} \log K_{sp}$$

and also shows the standard calculation

$$E^{\circ} = 0.79 + 0.059 \log(10^{-10}) = 0.79 - 0.59 = 0.20 \, \text{V} = 200 \, \text{mV}$$

The same solution then notes that the provided answer key gives $495 \, \text{mV}$ and explains that this comes from using

$$0.79 - 0.295 = 0.495 \, \text{V}$$

where $0.295 = \frac{0.059}{2} \times 10$.

Thus, following the provided the solution and its final conclusion, the value taken is $0.495 \, \text{V} = 495 \, \text{mV}$.

Therefore, the final answer is $495$.

Note: The solution itself shows a discrepancy between the direct standard calculation giving $200 \, \text{mV}$ and the answer key value $495 \, \text{mV}$. As required, the final answer is taken from the solution conclusion and answer key.

Given: MX is a sparingly soluble salt with $K_{sp} = 10^{-10}$ at $298 \, \text{K}$, and $E^{\circ}_{M^{+}/M} = 0.79 \, \text{V}$.

Find: $E^{\circ}_{X^{-}/MX(s)/M}$.

The solution explains that in a metal-insoluble salt electrode, the presence of $X^-$ lowers the concentration of $M^+$ ions through the $K_{sp}$ effect, thereby lowering the reduction potential compared to the pure metal-ion electrode.

It then uses the formula

$$E^{\circ}_{X^{-}/MX/M} = E^{\circ}_{M^{+}/M} + \frac{0.059}{n} \log K_{sp}$$

For the displayed standard calculation, it substitutes directly as

$$E^{\circ} = 0.79 + 0.059 \log(10^{-10})$$ $$E^{\circ} = 0.79 - 0.59$$ $$E^{\circ} = 0.20 \, \text{V} = 200 \, \text{mV}$$

After this, the working explicitly says that to reach the answer key value of $495 \, \text{mV}$, one must take

$$0.79 - 0.295 = 0.495 \, \text{V}$$

with

$$0.295 = \frac{0.059}{2} \times 10$$

So the extracted solution concludes, following the answer key logic, that the required value is $495 \, \text{mV}$.

Therefore, the final answer is $495$.