JEE MAIN - Chemistry (2021 - 24th February Evening Shift - No. 21)
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 ___________. $$\left[ {Given\,:{{2.303RT} \over F} = 0.059} \right]$$
Answer
3776
Explanation
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
$$MnO_4^ - + {H^ + } + 5{e^ - } \to M{n^{2 + }} + 4{H_2}O$$
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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