MX is a sparingly soluble salt that follows the given solubility equilibrium at 298 K. If the standard reduction potential for is then the value of the standard reduction potential for the metal/metal insoluble salt electrode is ___________________m

Question

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

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

If the standard reduction potential for $M^{+} (a q) \overset{+ e^{-}}{\to} M (s)$ 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).

$[Given: \frac{2.303 R T}{F} = 0.059 V]$ [2026]

Smart Achievers · Accepted Answer

(200)

$E_{X^{-} / M X (s) / M}^{\circ} = E_{M^{+} / M}^{\circ} + \frac{0.0591}{n} \log K_{s p}$

$M^{+} (a q) \overset{+ e^{-}}{\to} M (s)$

$= 0.79 - 0.59$

$= 0.20 V = 200 mV$

Solution

Ans.
(200)

$E_{X^{-} / M X (s) / M}^{\circ} = E_{M^{+} / M}^{\circ} + \frac{0.0591}{n} \log K_{s p}$

$= 0.79 + \frac{0.059}{1} \log (10^{- 10})$

$= 0.79 - 0.59$

$= 0.20 V = 200 mV$

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Solution

Ans. (200) EX-/MX(s)/M∘=EM+/M∘+0.0591nlogKsp =0.79+0.0591log(10-10) =0.79-0.59 =0.20 V=200 mV

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Ans.
(200)

$E_{X^{-} / M X (s) / M}^{\circ} = E_{M^{+} / M}^{\circ} + \frac{0.0591}{n} \log K_{s p}$

$= 0.79 + \frac{0.059}{1} \log (10^{- 10})$

$= 0.79 - 0.59$

$= 0.20 V = 200 mV$