MX is a sparingly soluble salt that follows the given solubility equilibrium at 298 K.
$$MX(s) \rightleftharpoons M^{+} (aq)+X^{-}(aq); K_{sp}=10^{-10}$$
If the standard reduction potential for $$M^{+}(aq)\xrightarrow {+e^{-}}M(s)$$ and $$\left(E_{M^{+}/M}^\ominus\right)=0.79V$$, then the value of the standard reduction potential for the metal/metal insoluble salt electrode $$E^{\ominus}_{X^{-}/MX (s)/M }$$ is __________mV.(nearest integer)
[Given : $$\frac{2.303RT}{F} = 0.059V$$]
Correct Answer: 200
Solution
Given $$K_{sp} = 10^{-10}$$ for MX, and $$E^\circ_{M^+/M} = 0.79$$ V.
The electrode reaction for metal/insoluble salt: $$MX(s) + e^- \to M(s) + X^-(aq)$$
This can be split as: $$MX(s) \to M^+(aq) + X^-(aq)$$ ($$K_{sp}$$) and $$M^+(aq) + e^- \to M(s)$$ ($$E^\circ$$).
$$E^\circ_{X^-/MX/M} = E^\circ_{M^+/M} + \frac{0.059}{1}\log K_{sp}$$
$$= 0.79 + 0.059 \times \log(10^{-10}) = 0.79 + 0.059 \times (-10) = 0.79 - 0.59 = 0.20$$ V = 200 mV
The answer is 200 mV.
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