JEE Main 2026 — Jan 21, Shift 2 — Chemistry solution

JEE Main2026Jan 21, Shift 2Chemistry

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

$M X (s) ⇌ M^{+} (a q) + X^{-} (a q); K_{s p} = 1 0^{- 10}$

If the standard reduction potential for $M^{+} (a q) + e^{-} \to M (s)$ is $(E_{M^{+} / M}^{⊖}) = 0.79$ V, then the value of the standard reduction potential for the metal/metal insoluble salt electrode $E_{X^{-} / M X (s) / M}^{⊖}$ is __________ mV. (nearest integer)

$[Given: \frac{2.303 R T}{F} = 0.059 V]$

Your answer

Solution

The two electrode half-reactions are related by adding the dissolution equilibrium of MX. Using the relation between the two standard potentials:

$E_{X^{-} / M X / M}^{⊖} = E_{M^{+} / M}^{⊖} + \frac{0.0591}{n} lo g K_{s p}$ .

Substitute $n = 1$ , $K_{s p} = 1 0^{- 10}$ : $E_{X^{-} / M X / M}^{⊖} = 0.79 + 0.059 \times (- 10) = 0.79 - 0.59 = 0.20$ V.

Converting to millivolts: 200 mV.

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