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JEE Main Chemistry Electrochemistry 2026 JEE Main 2026 (21 January Shift 2)

JEE Main Chemistry Question (2026) — Solution

Question

MX is a sparingly soluble salt that follows the given solubility equilibrium at 298 K. MX ( s ) M ^ + ( aq )+ X ^ - ( aq ) ; K _ sp =10^ -10 If the standard reduction potential for M ^ + ( aq ) + e ^ - 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 ^ - / MX ( s ) / M ^ is \_\_\_\_ mV. (nearest integer) [Given : 2.303 RT F =0.059 ~V ]

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

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 of the salt. The half-cell reaction for the X^-/MX(s)/M electrode is: MX(s) + e^- M(s) + X^-(aq). This can be viewed as a combination of two processes: 1. MX(s) M^+(aq) + X^-(aq) with equilibrium constant K_ sp = 10^ -10 2. M^+(aq) + e^- M(s) with E^ _ M^+/M = 0.79 V The relation between the standard potentials is given by: E^ _ X^-/MX/M = E^ _ M^+/M + 2.303RT nF _ 10 K_ sp Given n = 1 and 2.303RT F = 0.059 V : E^ _ X^-/MX/M = 0.79 + 0.059 _ 10 (10^ -10 ) E^ _ X^-/MX/M = 0.79 + 0.059(-10) E^ _ X^-/MX/M = 0.79 - 0.59 = 0.20 V To convert the value into mV: 0.20 V = 0.20 1000 mV = 200 mV

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