The equilibrium constant of a 2 electron redox reaction at 298 K is 3.8 \times 10^{–3}. The cell potential Eº (in V) and the free energy change ΔGº (in kJ mol^{–1}) for this equilibrium respectively, are

Correct Answer is : – 0.071, 13.8

KVPY SX SB Exam 2016 Question Paper

Question : 57 of 120

Marks: +1, -0

The equilibrium constant of a 2 electron redox reaction at 298 K is

3.8 × 10^{–3}

. The cell potential Eº (in V) and the free energy change ΔGº (in kJ mol

^{–1}

) for this equilibrium respectively, are

Solution:

{K}_{ {eq}}=3.8 × 10^{-3}

{n}=

no. of electrons involved in the reaction

=2

Now,

∆ {G}^{∘}=- {RT} \ln {K}

=-8.314 × 298 × \ln (3.8 × 10^{-3})

=13806 {Jmol}^{-1}=13.8 {kJmol}^{-1}

Now,

∆ {G}^{∘}=- {nFE}^{∘}

∴ {E}^{∘}=-\;{∆ {G}^{∘}}/{ {nF}}

=-\;{3.8 × 10^{3}}/{2 × 96500}=-0.071 {V}

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