Given below are the half-cell reactions- Mn2- - 2e- - Mn- Eo - -1-18V 2Mn3- - e- - Mn2- Eo - -1-51VThe Eo for 3Mn2- Mn -2Mn3- will be-

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Redox Reactions and Couple

Given below a...

Question

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:

- 0.33 V

; the reaction will not occur

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- 0.33 V

; the reaction will occur

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- 2.69 V

; the reaction will not occur

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- 2.69 V

; the reaction will occur

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Solution

The correct option is B $- 2.69 V$ ; the reaction will not occur
Standard electrode potential of reaction will not change due to multiply the half-cell reactions with some numbers,
To get the main eq we have to reverse $2 n d$ equation and add them
So $E_{3} = E_{2} + E_{1}$
$E_{3} = - 1.18 + (- 1.51)$
$E_{3} = - 2.69 V$
The reaction is not possible as the $Δ G$ will come +ve for this case and that indicates reaction is non-spontaneous.

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Similar questions

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:

Given below are the half-cell reaction:
$M n^{2 +} + 2 e^{⊖} \to M n; E^{⊖} = - 1.18 V$
$2 (M n^{3 +} + e^{⊖} \to M n^{2 +}; E^{⊖} = + 1.51 V$

The $E^{⊖}$ for $3 M n^{2 +} \to M n + 2 M n^{3 +}$ will be:
(IIT-JEE 2014)

Q. Given below are the half-cell reaction:

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

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

The

E^{⊖}

for

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

will be:
(IIT-JEE 2014)

Q. Given below are the half-cell reactions:

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

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

The

E^{\circ}

for

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

will be:

Q. Consider the half-cell reduction reactions :

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

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

The

E^{o}

for the reaction

{3 M n}^{2 +} \to {M n}^{0} + {2 M n}^{3 +}

and possibility of the forward reaction are respectively:

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Given below are the half-cell reactions:Mn2++2e−→Mn;Eo=−1.18V2(Mn3++e−→Mn2+);Eo=+1.51VThe Eo for 3Mn2+→Mn+2Mn3+ will be:

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: