The Eo for- Mn4- - 2e - Mn2- is equal to-

The correct option is D 1.23 VAccording to given values, Mn^3+ + e → Mn^2+ #160; #160; #160; #160; #160; #160; #160; Δ G_1^o = 1 × 1.5 ...

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Standard XII

Chemistry

Nernst Equation

The Eo for,...

Question

The $E^{o}$ for, $M n^{4 +} + 2 e \to M n^{2 +}$ is equal to:

1.23 V

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2.46 V

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0.56 V

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-0.56 V

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Solution

The correct option is D 1.23 V
According to given values,
$M n^{3 +} + e \to M n^{2 +} - Δ G_{1}^{o} = 1 \times 1.51 \times F$
$M n^{4 +} + e \to M n^{3 +} - Δ G_{2}^{o} = 1 \times 0.95 \times F$
by adding these equations,
$M n^{4 +} + 2 e \to M n^{2 +}; - (Δ G_{1}^{o} + Δ G_{2}^{o}) = (1.51 + 0.95) \times F$
So for this equation,
$- Δ G_{3}^{o} = - (Δ G_{1}^{o} + Δ G_{2}^{o}) (Δ G = n F E^{o})$
$2 \times E_{3}^{o} \times F = 2.46 \times F$
$E_{3}^{o} = 1.23 V$

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Similar questions

E_{R P}^{o}

and their respective half reactions for some changes are given below:

(a)

M n O_{4}^{-} (a q) + 8 H^{+} + 5 e \to M n^{2 +} (a q) + 4 H_{2} O (l); E^{o} = + 1.52 V

(b)

M n O_{4}^{2 -} (a q) + 4 H^{+} + 2 e \to M n O_{2} (s) + 2 H_{2} O (l); E^{o} = + 2.26 V

(c)

2 M n O_{4}^{-} (a q) + 2 e \to 2 M n O_{4}^{2 -} (a q); E^{o} = + 0.56 V

(acid med.)
(d)

M n^{3 +} + e \to M n^{2 +}; E^{o} = + 1.51 V

(acid med.)
(e)

M n^{4 +} + e \to M n^{3 +}; E^{o} = 0.95 V

(acid med.)

The most stable oxidation state of Mn is :

Q. Given below are the half-cell reactions

M n^{2 +} + 2 e^{-} \to M n, E^{o} = - 1.18 V

2 (M n^{3 +} + e^{-} \to M n^{2 +}), E^{o} = + 1.51 V

The

E^{o}

for

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

will be:

Q. Given below are the half-cell reactions:

M n^{2 +} + 2 e^{-} \to M n; E^{o} = - 1.18 V

2 (M n^{3 +} + e^{-} \to M n^{2 +}); E^{o} = + 1.51 V

The

E^{o}

for

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

will be:

Q. Given,

C u^{2 +} + 2 e^{-} \to C u; E^{o} = 0.337

C u^{2 +} + e^{-} \to C u^{+}; E^{o} = - 0.153

E^{o}

value for the reaction,

C u^{+} + e \to C u

will be:

Q. For the reaction,

F e^{3 +} + 3 e ⇌ F e, E^{o}

is -0.036 V and the standard electrode potential(SEP) for

F e^{3 +} + e ⇌ F e^{2 +}

is 0.77 V.
Hence,

E^{o}

for

F e^{2 +} + 2 e ⇌ F e

is:

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The Eo for, Mn4++2e→Mn2+ is equal to:

The correct option is D 1.23 VAccording to given values,Mn3++e→Mn2+−ΔGo1=1×1.51×FMn4++e→Mn3+−ΔGo2=1×0.95×Fby adding these equations,Mn4++2e→Mn2+;−(ΔGo1+ΔGo2)=(1.51+0.95)×FSo for this equation,−ΔGo3=−(ΔGo1+ΔGo2)(ΔG=nFEo)2×Eo3×F=2.46×FEo3=1.23V

The $E^{o}$ for, $M n^{4 +} + 2 e \to M n^{2 +}$ is equal to: