Chemistry · 2026
JEE Main · 21 January 2026, Shift 2 · Q74
MX is a sparingly soluble salt that follows the given solubility equilibrium at 298 K. MX ( s ) ⇌ M^+( aq )+ X^-( aq ); K_sp =10^-10 If the standard…
MX is a sparingly soluble salt that follows the given solubility equilibrium at $\displaystyle 298$ 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 $\displaystyle \mathrm{M}^{+}(\mathrm{aq}) \xrightarrow{+\mathrm{e}^{-}} \mathrm{M}(\mathrm{s})$ is $\displaystyle \left(\mathrm{E}_{\mathrm{M}^{+} / \mathrm{M}}^{\ominus}\right)=0.79 \mathrm{~V}$, then the value of the standard reduction potential for the metal/metal insoluble salt electrode $\displaystyle \mathrm{E}_{\mathrm{X}^{-} / \mathrm{MX}(\mathrm{s}) / \mathrm{M}}^{\ominus}$ is
$\displaystyle \_\_\_\_$ mV. (nearest integer)
[Given : $\displaystyle \frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{~V}$ ]
Official answer
From NTA’s final answer key for this paper.
200
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JEE Main 2026 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.