MX is a sparingly soluble salt that follows the given solubility equilibrium at 298 K $\operatorname{MX}(\mathrm{s}) \rightleftharpoons \mathrm{M}^{+}(\mathrm{aq})+\mathrm{X}^{-}(\mathrm{aq})$; $\mathrm{K}_{\mathrm{sp}}=10^{-10}$ If the standard reduction potential for $\mathrm{M}^{+}(\mathrm{aq}) \xrightarrow{+\mathrm{e}^{-}} \mathrm{M}(\mathrm{s})$ is $(\mathrm{E}_{\mathrm{M}^{+} / \mathrm{M}}^{\ominus})=0.79 \mathrm{~V}$, then the value of the standard reduction potential for the metal/metal insoluble salt electrode $\mathrm{E}_{\mathrm{X}^{-} / \mathrm{MX}(\mathrm{s}) / \mathrm{M}}^{\Theta}$ is $\_\_\_\_$ mV. (nearest integer) [Given: $\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{~V}$]

JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Redox Reactions and Electrochemistry

MX is a sparingly soluble salt that follows the given solubility equilibrium at 298 K

MX (s) ⇌ M^{+} (aq) + X^{-} (aq)

;

K_{sp} = 1 0^{- 10}

If the standard reduction potential for

M^{+} (aq) + e^{-} M (s)

(E_{M^{+} / M}^{⊖}) = 0.79 V

, then the value of the standard reduction potential for the metal/metal insoluble salt electrode

E_{X^{-} / MX (s) / M}^{Θ}

____

mV. (nearest integer)
[Given:

\frac{2.303 RT}{F} = 0.059 V

]

Answer: 200

Combining the dissolution

MX (s) ⇌ M^{+} + X^{-}

of solubility product

K_{s p}

with the reduction

M^{+} + e^{-} \to M

gives the electrode reaction

MX (s) + e^{-} \to M + X^{-}

, whose standard potential follows as

E_{X^{-} / MX/M}^{\circ} = E_{M^{+} / M}^{\circ} + 0.059 lo g K_{s p}

for the one-electron change. Substituting,

E^{\circ} = 0.79 + 0.059 lo g (1 0^{- 10}) = 0.79 - 0.59 = 0.20 V

. The standard reduction potential of the metal/insoluble-salt electrode is 200 mV.

JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Integer answer