Q21 / 75 · JEE Main 2026 (21 January Shift 2) · Chemistry
MX is a sparingly soluble salt that follows the given solubility equilibrium at 298 K.
$\mathrm{MX}(\mathrm{s}) \rightleftharpoons \mathrm{M}^{+}(\mathrm{aq})+\mathrm{X}^{-}(\mathrm{aq}) ; \mathrm{K}_{\mathrm{sp}}=10^{-10}$
If the standard reduction potential for $\mathrm{M}^{+}(\mathrm{aq}) \xrightarrow{+\mathrm{e}^{-}} \mathrm{M}(\mathrm{s})$ is $\left(\mathrm{E}_{\mathrm{M}^{+} / \mathrm{M}}^{\ominus}\right)=0.79 \mathrm{~V}$, then the value of the standard reduction potential for the metal/metal insoluble salt electrode $\mathrm{E}_{\mathrm{X}^{-} / \mathrm{MX}(\mathrm{s}) / \mathrm{M}}^{\ominus}$ is $\_\_\_\_$ mV. (nearest integer)
[Given : $\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{~V}$ ]
$\mathrm{MX}(\mathrm{s}) \rightleftharpoons \mathrm{M}^{+}(\mathrm{aq})+\mathrm{X}^{-}(\mathrm{aq}) ; \mathrm{K}_{\mathrm{sp}}=10^{-10}$
If the standard reduction potential for $\mathrm{M}^{+}(\mathrm{aq}) \xrightarrow{+\mathrm{e}^{-}} \mathrm{M}(\mathrm{s})$ is $\left(\mathrm{E}_{\mathrm{M}^{+} / \mathrm{M}}^{\ominus}\right)=0.79 \mathrm{~V}$, then the value of the standard reduction potential for the metal/metal insoluble salt electrode $\mathrm{E}_{\mathrm{X}^{-} / \mathrm{MX}(\mathrm{s}) / \mathrm{M}}^{\ominus}$ is $\_\_\_\_$ mV. (nearest integer)
[Given : $\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{~V}$ ]
Correct answer: 200
Solution
The standard reduction potential for a metal/metal insoluble salt electrode is related to the standard reduction potential of the metal ion and the solubility product $K_{sp}$ of the salt.
The half-cell reaction for the $X^-/MX(s)/M$ electrode is: $MX(s) + e^- \rightarrow M(s) + X^-(aq)$.
This can be viewed as a combination of two processes:
1. $MX(s) \rightleftharpoons M^+(aq) + X^-(aq)$ with equilibrium constant $K_{sp} = 10^{-10}$
2. $M^+(aq) + e^- \rightarrow M(s)$ with $E^\ominus_{M^+/M} = 0.79\text{ V}$
The relation between the standard potentials is given by:
$E^\ominus_{X^-/MX/M} = E^\ominus_{M^+/M} + \frac{2.303RT}{nF} \log_{10} K_{sp}$
Given $n = 1$ and $\frac{2.303RT}{F} = 0.059\text{ V}$:
$E^\ominus_{X^-/MX/M} = 0.79 + 0.059 \log_{10}(10^{-10})$
$E^\ominus_{X^-/MX/M} = 0.79 + 0.059(-10)$
$E^\ominus_{X^-/MX/M} = 0.79 - 0.59 = 0.20\text{ V}$
To convert the value into mV:
$0.20\text{ V} = 0.20 \times 1000\text{ mV} = 200\text{ mV}$
The half-cell reaction for the $X^-/MX(s)/M$ electrode is: $MX(s) + e^- \rightarrow M(s) + X^-(aq)$.
This can be viewed as a combination of two processes:
1. $MX(s) \rightleftharpoons M^+(aq) + X^-(aq)$ with equilibrium constant $K_{sp} = 10^{-10}$
2. $M^+(aq) + e^- \rightarrow M(s)$ with $E^\ominus_{M^+/M} = 0.79\text{ V}$
The relation between the standard potentials is given by:
$E^\ominus_{X^-/MX/M} = E^\ominus_{M^+/M} + \frac{2.303RT}{nF} \log_{10} K_{sp}$
Given $n = 1$ and $\frac{2.303RT}{F} = 0.059\text{ V}$:
$E^\ominus_{X^-/MX/M} = 0.79 + 0.059 \log_{10}(10^{-10})$
$E^\ominus_{X^-/MX/M} = 0.79 + 0.059(-10)$
$E^\ominus_{X^-/MX/M} = 0.79 - 0.59 = 0.20\text{ V}$
To convert the value into mV:
$0.20\text{ V} = 0.20 \times 1000\text{ mV} = 200\text{ mV}$
Hence, the correct answer is 200.