Calculate the standard electrode potential for the electrode MnO 4- MnO 2 in an acid solution- E -circ- Mn - O - 4- Mn -2-1-51 - V -E - MnO - 2- Mn -2-1-23 - VA- 1-51 VB- 1-70 VC- -1-51 VD- 2-74 V

The correct option is B 1.70 VMnO^ _4 + 8H^+ 5e^ → Mn^2+ + 4 H_2 O E^∘ = 1.5 V ∴Δ G^∘ = 5(1.51)F = 7.55 F Mn^ _2 + 4H^+ 2e^ → mn^2+ + 2H_2O E^∘ = 1.2 ...

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Standard XII

Chemistry

Redox Reactions

Calculate the...

Question

Calculate the standard electrode potential for the electrode $\frac{M n O_{4}^{-}}{M n O_{2}}$ in an acid solution?
$E^{\circ} M n_{4}^{O -}, M n^{2 +} = 1.51 V E^{\circ} M n O_{2}^{-}, M n^{2 +} = 1.23 V$

1.51 V

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1.70 V

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-1.51 V

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2.74 V

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Solution

The correct option is B 1.70 V
$M n O_{4}^{-} + 8 H^{+} 5 e^{-} \to M n^{2 +} + 4 H_{2} O E^{\circ} = 1.5 V ∴ Δ G^{\circ} = - 5 (1.51) F = - 7.55 F M n_{2}^{-} + 4 H^{+} 2 e^{-} \to m n^{2 +} + 2 H_{2} O E^{\circ} = 1.23 V ∴ Δ G^{\circ} = - 2 (1.23) F = - 2.46 F M n O_{4}^{-} - M n O_{2} + 4 H^{f} + 3 e \to 2 H_{2} O Δ G^{\circ} = - 7.55 F - (- 2.46 F)$
On subtracting
$= - 5.09 F ∴ E^{\circ} M n O_{4}^{-}, M n O_{2} = \frac{Δ G^{\circ}}{- n F} = \frac{- 5.09 F}{- 3 F} = 1.70 V o l t$

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Similar questions

Q. What is the standard electrode potential for the electrode,

M n O_{4}^{-} | M n O_{2}

in an acid solution?

E_{M n O_{4}^{-} / M n^{2 +}}^{0} = 1.51 V E_{M n O_{2} / M n^{2 +}}^{0} = 1.23 V

Q. The standard electrode potential in V for the electrode

M n O_{4}^{-} / M n O_{2}

in solution is: (round off the answer to the nearest integer)
Given:

E_{M n O_{4}^{-} / M n^{2 +}}^{o} = 1.51 V

and

E_{M n O_{2} / M n^{2 +}}^{o} = 1.23 V

Q. Find the magnitude of the standard electrode potential of

M n O_{4}^{-} / M n O_{2}

. The standard electrode potentials of

M n O_{4}^{-} / M n^{2 +} = 1.51 V

and

M n O_{2} / M n^{2 +} = 1.23 V

Q. What is the standard electrode potential for the electrode

M n O_{4}^{-} / M n O_{2}

in solution?
(Given:

E_{M n O_{4}^{-} / M n^{2 +}}^{\circ} = 1.51 v o l t, E_{M n O_{2} / M n^{2 +}}^{\circ} = 1.23 v o l t

Q. What is the standard electrode potential for the electrode

M n O {_{4}}^{-} / M n O_{2}

in the solution?
Given:

E {^{o}}_{M n O {_{4}}^{-} / M n^{2 +}} = 1.51 V

and

E {^{o}}_{M n O_{2} / M n^{2 +}} = 1.23 V

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Calculate the standard electrode potential for the electrode MnO−4MnO2 in an acid solution? E∘ MnO−4,Mn2+=1.51 VE∘ MnO−2,Mn2+=1.23 V

The correct option is B 1.70 VMnO−4+8H+5e−→Mn2++4H2OE∘=1.5V∴ΔG∘=−5(1.51)F=−7.55FMn−2+4H+2e−→mn2++2H2OE∘=1.23V∴ΔG∘=−2(1.23)F=−2.46FMnO−4−MnO2+4Hf+3e→2H2OΔG∘=−7.55F−(−2.46F) On subtracting =−5.09F∴E∘MnO−4,MnO2=ΔG∘−nF=−5.09F−3F=1.70Volt

Calculate the standard electrode potential for the electrode $\frac{M n O_{4}^{-}}{M n O_{2}}$ in an acid solution?
$E^{\circ} M n_{4}^{O -}, M n^{2 +} = 1.51 V E^{\circ} M n O_{2}^{-}, M n^{2 +} = 1.23 V$