The standard potential is given below- of the following at 25- C- MnO 2- Mn 3- E -0-95 VMn -3- Mn 2- E -1-51 VThe standard potential of MnO 2- Mn 2- is -A- -2-46 VB- -1-23 VC- -0-56 VD- 1-23 V

The correct option is D 1.23V2e^ +MnO_2 → Mn^2+, E^∘=? Δ G= 2FE^∘ (i) e^ +MnO_2→ Mn^3+Δ G^0_1= FE^0= 0.95V (ii) e^ +Mn^3+→ Mn^2+Δ G^0_2= FE^0= 1.51V Equat ...

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Standard XII

Chemistry

Gibb's Free Energy

The standard ...

Question

The standard potential is given below, of the following at $25^{\circ} C$ .
$M n O_{2} \to M n^{3 +} E^{\circ} = 0.95 V$
$M n^{+ 3} \to M n^{2 +} E^{\circ} = 1.51 V$
The standard potential of $M n O_{2} \to M n^{2 +}$ is :

- 0.56 V

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- 2.46 V

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- 1.23 V

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1.23 V

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Solution

The correct option is D $1.23 V$
$2 e^{-} + M n O_{2} \to M n^{2 +}, E^{\circ} = ?$
$Δ G = - 2 F E^{\circ}$
$(i) e^{-} + M n O_{2} \to M n^{3 +} Δ G_{1}^{0} = - F E^{0} = - 0.95 V$
$(i i) e^{-} + M n^{3 +} \to M n^{2 +} Δ G_{2}^{0} = - F E^{0} = - 1.51 V$
Equation (i) + (ii) give the desired equations
$∴ - 2 F E^{0} = (- 0.95 - 1.51) F$
$E^{0} = 1.23 V$

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Similar questions

Q. The standard potential is given below, of the following at

25^{\circ} C

M n O_{2} \to M n^{3 +} E^{\circ} = 0.95 V

M n^{+ 3} \to M n^{2 +} E^{\circ} = 1.51 V

The standard potential of

M n O_{2} \to M n^{2 +}

is :

Q. Given the standard potential of the following at

25^{\circ} C

M n O_{2} \to M n^{3 +}

;

E^{⊖} = 0.95 V

M n^{3 +} \to M n^{2 +}

;

E^{⊖} = 1.51 V

The standard potential of

M n O_{2} \to M n^{2 +}

is:

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 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

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

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The standard potential is given below, of the following at 25∘C. MnO2→Mn3+ E∘=0.95V Mn+3→Mn2+ E∘=1.51V The standard potential of MnO2→Mn2+ is :

The correct option is D 1.23V2e−+MnO2→Mn2+,E∘=? ΔG=−2FE∘ (i) e−+MnO2→Mn3+ΔG01=−FE0=−0.95V (ii) e−+Mn3+→Mn2+ΔG02=−FE0=−1.51V Equation (i) + (ii) give the desired equations ∴−2FE0=(−0.95−1.51)F E0=1.23V

The standard potential is given below, of the following at $25^{\circ} C$ .
$M n O_{2} \to M n^{3 +} E^{\circ} = 0.95 V$
$M n^{+ 3} \to M n^{2 +} E^{\circ} = 1.51 V$
The standard potential of $M n O_{2} \to M n^{2 +}$ is :