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← JEE Main 2026 · 4 Apr · Shift 2
ChemistryInteger+4 / -1Question 73 of 75

JEE Main 2026 · 4 Apr · Shift 2, Chemistry question 73

An electrochemical cell, consist of the following two redox couples, MXx+\ce{M^{x+}}(aq)/M(s) [Ered⊖=+0.15 V][E^\ominus_{red}=+0.15\ \mathrm V] and FeX3+\ce{Fe^{3+}}(aq)/Fe(s) [Ered⊖=−0.036 V][E^\ominus_{red}=-0.036\ \mathrm V]. The cell EMF (EcellE_\text{cell}) is recorded to be 0.2057 V. If the reaction quotient of the electrochemical reaction is found to be 10−210^{-2}, then the value of xx is ________. (Nearest integer)
[Given : M is a p-block metal and 2.303RTF=0.059\dfrac{2.303RT}F=0.059 V]

Answer and worked solution

Official answer: 2

Cell. M couple (+0.15 V) is the cathode and Fe (−0.036 V) the anode, so Ecell∘=0.186E^\circ_\text{cell}=0.186 V. The reaction is 3 MXx++x Fe→3 M+x FeX3+\ce{3M^{x+} + xFe -> 3M + xFe^{3+}} with n=3xn=3x.

Nernst. 0.2057=0.186−0.059nlog⁡10−20.2057=0.186-\frac{0.059}n\log10^{-2} gives 0.118n=0.0197\frac{0.118}n=0.0197, so n=6n=6 and x=2x=2.

Answer: 2