Q.75. MX is a sparingly soluble salt that follows the given solubility equilibrium at 298 K MX(s) M

Q: What is the correct answer for JEE Main Chemistry Q75?

The correct answer is 200. Sol. For the reaction MX(s) + e⁻ $\rightleftharpoons$ M(s) + X⁻(aq), the Nernst equation is $E = E^{\circ} - (0.059/n) \log Q$. At equilibrium, $E = E^{\circ}_{M+/M} - (0.059/1) \log(1/[M^+])$. We need $E^{\circ}_{X⁻/MX(s)/M}$. This corresponds to the half-reaction MX(s) + e⁻ $\rightleftharpoons$ M(s) + X⁻. The standard potential for this is related to K$_{sp}$. We have $E^{\circ}_{X⁻/MX(s)/M} = E

Q: Q.75. MX is a sparingly soluble salt that follows the given solubility equilibrium at 298 K MX(s) M⁺(aq) + X⁻(aq); K = 10⁻¹⁰ If the standard reduction potential for M⁺(aq) + e⁻ → M(s) is (E°) =…

The correct answer is 200. Sol. For the reaction MX(s) + e⁻ $\rightleftharpoons$ M(s) + X⁻(aq), the Nernst equation is $E = E^{\circ} - (0.059/n) \log Q$. At equilibrium, $E = E^{\circ}_{M+/M} - (0.059/1) \log(1/[M^+])$. We need $E^{\circ}_{X⁻/MX(s)/M}$. This corresponds to the half-reaction MX(s) + e⁻ $\rightleftharpoons$ M(s) + X⁻. The standard potential for this is related to K$_{sp}$. We have $E^{\circ}_{X⁻/MX(s)/M} = E

Question

Q.75. MX is a sparingly soluble salt that follows the given solubility equilibrium at 298 K MX(s)  M⁺(aq) + X⁻(aq); K = 10⁻¹⁰ If the standard reduction potential for M⁺(aq) + e⁻ → M(s) is (E°) = 0.79V, then the value of the standard reduction potential for the metal/metal insoluble salt electrode E° is  mV. (nearest integer) [Given: ]

FirstInTest · Accepted Answer

Sol. For the reaction MX(s) + e⁻ $ightleftharpoons$ M(s) + X⁻(aq), the Nernst equation is $E = E^{\circ} - (0.059/n) \log Q$. At equilibrium, $E = E^{\circ}_{M+/M} - (0.059/1) \log(1/[M^+])$. We need $E^{\circ}_{X⁻/MX(s)/M}$. This corresponds to the half-reaction MX(s) + e⁻ $ightleftharpoons$ M(s) + X⁻. The standard potential for this is related to K$_{sp}$. We have $E^{\circ}_{X⁻/MX(s)/M} = E^{\circ}_{M+/M} + (0.059/n) \log K_{sp}$. Here $n=1$. $E^{\circ}_{X⁻/MX(s)/M} = 0.79 + (0.059/1) \log(10^{-10}) = 0.79 + 0.059 	imes (-10) = 0.79 - 0.59 = 0.20$ V. In mV, $0.20 	imes 1000 = 200$ mV.

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