MnO-4 -8H - - 5e- - Mn2- -4H2O -EO-1-51 V - - G01-5- 1-51 - FMnO2 -4H - - 2e- - Mn2- -2H2O -EO-1-23 V - - G02-5- 1-23- FE- MnO4-MnO2 is

MnO^ _4+8H^++5e^ → Mn^2++4H_2O E^∘ =1.51 V ....(1) Δ G_1^∘ MnO_2+4H^++2e^ → Mn^2++2H_2O E^∘ =1.23 V ....(2) Δ G_2^∘ On reversing equation (2) we get nbsp; ...

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Byju's Answer

Standard XII

Chemistry

Common Oxidation States

MnO-4 +8H + +...

Question

$M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O; E^{O} = 1.51 V; Δ G_{1}^{0} = - 5 \times 1.51 \times F$
$M n O_{2} + 4 H^{+} + 2 e^{-} \to M n^{2 +} + 2 H_{2} O; E^{O} = 1.23 V; Δ G_{2}^{0} = - 5 \times 1.23 \times F$
$E_{M n O_{4}^{-} | M n O_{2}}^{\circ}$ is

1.7 V

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0.91 V

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1.37 V

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0.55 V

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Solution

The correct option is A $1.7 V$
$M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$
$E^{\circ} = 1.51 V . . . . (1) Δ G_{1}^{\circ}$
$M n O_{2} + 4 H^{+} + 2 e^{-} \to M n^{2 +} + 2 H_{2} O$
$E^{\circ} = 1.23 V . . . . (2) Δ G_{2}^{\circ}$
On reversing equation $(2)$ we get
$M n^{2 +} + 2 H_{2} O \to M n O_{2} + 4 H^{+} + 2 e^{-} . . . (3)$
$E^{\circ} = - 1.23 V . . . (3) Δ G_{3}^{\circ}$
On addition of equation $(1), (3)$ we get,
$M n O_{4}^{-} + 4 H^{+} + 3 e^{-} \to 2 H_{2} O + M n O_{2}$
$E_{M n O_{4}^{-} | M n O_{2}}^{\circ} Δ G_{4}^{0}$
$We know, Δ G^{0} = - n F E^{\circ}$
$So, Δ G_{4}^{0} = Δ G_{1}^{\circ} + Δ G_{3}^{\circ}$
$\Rightarrow - 3 F E_{M n O_{4}^{-} / M n O_{2}}^{0} = - 5 F (1.51) + [- 2 F (- 1.23)]$
$\Rightarrow - 3 E_{M n O_{4}^{-} / M n O_{2}}^{0} = - 5 \times 1.51 + 2 \times 1.23$
$\Rightarrow - 3 E_{M n O_{4}^{-} / M n O_{2}}^{0} = - 7.55 + 2.46$
$\Rightarrow E_{M n O_{4}^{-} / M n O_{2}}^{0} = \frac{5.09}{3} = 1.696 \approx 1.7 V$

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Similar questions

Q. Calculate the value of

E_{M n O_{4}^{-} | M n O_{2}}^{\circ}

M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O

;

E^{\circ} = 1.51 V

M n O_{2} + 4 H^{+} + 2 e^{-} \to M n^{2 +} + 2 H_{2} O

;

E^{\circ} = 1.23 V

Q. The following two redox reactions are given:
i)

M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O; E^{o} = X_{1} V

ii)

M n O_{2} + 4 H^{+} + 2 e^{-} \to M n^{2 +} + 2 H_{2} O; E^{o} = X_{2} V

Find

E^{o}

for the following reaction.

M n O_{4}^{-} + 4 H^{+} + 3 e^{-} \to M n O_{2} + 2 H_{2} O

Q. For the following half cell reactions,

E^{o}

values are also given:

M n^{2 +} + 2 H_{2} O \to M n O_{2} + 4 H^{+} + 2 e^{-}

;

E^{o} = - 1.23 V

M n O_{4}^{-} + 4 H^{+} + 3 e^{-} \to M n O_{2} + 2 H_{2} O

;

E^{o} = + 1.70 V

Select the correct statements.

$M n {O_{4}}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O E^{o} = 1.51 V$
The quantity of electricity required in Faraday to reduce five moles of $M n O_{4}^{-}$ is ____.