JEE Main 2026 — Chemistry Physical Chemistry

The standard reduction potential for a metal/metal insoluble salt electrode is related to the standard reduction potential of the metal ion and the solubility product $K_{sp}$Ksp​ of the salt.

The half-cell reaction for the $X^-/MX(s)/M$X−/MX(s)/M electrode is: $MX(s) + e^- \rightarrow M(s) + X^-(aq)$MX(s)+e−→M(s)+X−(aq).

This can be viewed as a combination of two processes:

1. $MX(s) \rightleftharpoons M^+(aq) + X^-(aq)$MX(s)⇌M+(aq)+X−(aq) with equilibrium constant $K_{sp} = 10^{-10}$Ksp​=10−10

2. $M^+(aq) + e^- \rightarrow M(s)$M+(aq)+e−→M(s) with $E^\ominus_{M^+/M} = 0.79\text{ V}$EM+/M⊖​=0.79 V

The relation between the standard potentials is given by:

$$E^\ominus_{X^-/MX/M} = E^\ominus_{M^+/M} + \frac{2.303RT}{nF} \log_{10} K_{sp}$$EX−/MX/M⊖​=EM+/M⊖​+nF2.303RT​log10​Ksp​

Given $n = 1$n=1 and $\frac{2.303RT}{F} = 0.059\text{ V}$F2.303RT​=0.059 V:

$$E^\ominus_{X^-/MX/M} = 0.79 + 0.059 \log_{10}(10^{-10})$$EX−/MX/M⊖​=0.79+0.059log10​(10−10)

$$E^\ominus_{X^-/MX/M} = 0.79 + 0.059(-10)$$EX−/MX/M⊖​=0.79+0.059(−10)

$$E^\ominus_{X^-/MX/M} = 0.79 - 0.59 = 0.20\text{ V}$$EX−/MX/M⊖​=0.79−0.59=0.20 V

To convert the value into mV:

$$0.20\text{ V} = 0.20 \times 1000\text{ mV} = 200\text{ mV}$$0.20 V=0.20×1000 mV=200 mV