QJEEM21Ch5fae: JEE Main 2021 Chemistry - Class 12 JEEM21Ch5fae Solution: The magnitude of the change in oxidising power of the couple is …

Question

Verified Solution • 2025–26 Session moderate

The magnitude of the change in oxidising power of the

MnO_4^ - /M{n^{2 + }}

couple is x

\times

10^{$-$ 4} V, if the H⁺ concentration is decreased from 1M to 10^{$-$ 4} M at 25

^\circ

C. (Assume concentration of

MnO_4^ -

and

M{n^{2 + }}

to be same on change in H⁺ concentration). The value of x is ___________.

\left[ {Given\,:{{2.303RT} \over F} = 0.059} \right]

Accepted Answer

Reaction,

MnO_4^ - + {H^ + } + 5{e^ - } \to M{n^{2 + }} + 4{H_2}O

n = 5

Applying Nernst equation,

{E_{cell}} = E_{cell}^o - {{0.0591} \over n}\log {{[P]} \over {[R]}}

or

{E_{cell}} = E_{cell}^o - {{0.0591} \over n}\log {{[M{n^{2 + }}]} \over {[MnO_4^ - ]}}{\left[ {{1 \over {{H^ + }}}} \right]^8}

(I) Given, [H⁺] = 1 M

{E_1} = E^\circ - {{0.0591} \over 5}\log {{[M{n^{2 + }}]} \over {[MnO_4^ - ]}}

(II) Now, [H⁺] = 10^{$-$ 4} M

{E_2} = E^\circ - {{0.0591} \over 5}\log {{[M{n^{2 + }}]} \over {[MnO_4^ - ]}} \times {1 \over {{{({{10}^{ - 4}})}^8}}}

\therefore

\left| {{E_1} - {E_2}} \right|

\left| {{E_1} - {E_2}} \right| = {{0.0591} \over 5} \times 32 = 0.3776\,V = 3776 \times {10^{ - 4}}

x = 3776

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