Given below are the half-cell reactions- Mn2-2e- Mn- E0-1-18V 2 Mn3-e- Mn2- - E0-1-51V-The E0 for 3Mn2- Mn-2Mn3- will be-

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Given below a...

Question

Given below are the half-cell reactions:
$M n^{2 +} + 2 e^{-} \to M n; E^{0} = - 1.18 V$
$2 (M n^{3 +} + e^{-} \to M n^{2 +}); E^{0} = + 1.51 V$ .

The $E^{0}$ 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 D $- 2.69 V;$ the reaction will not occur
Adding equation $2$ from $1$ gives $3 {M n}^{2 +} \to M n + 2 {M n}^{3 +}$

Hence, $E^{0}$ for this reaction $= (- 1.18) - (+ 1.51) V$
$= - 2.69 V$ .

The reaction will not occur as $E_{c e l l}^{0}$ is negative value as for spontaneous reaction value should be positive.

Option C is correct.

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Similar questions

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

M n^{2 +} + 2^{e -} \to M n; E^{0} = - 1.18 V

2 (M n^{3 +} + e^{-} \to M n^{2 +}); E^{0} = + 1.51 V

The

E^{0}

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 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:

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Given below are the half-cell reactions: Mn2++2e−→Mn;E0=−1.18V 2(Mn3++e−→Mn2+);E0=+1.51V.The E0 for 3Mn2+→Mn+2Mn3+ will be:

The correct option is D −2.69V; the reaction will not occurAdding equation 2 from 1 gives 3Mn2+→Mn+2Mn3+Hence, E0 for this reaction =(−1.18)−(+1.51) V=−2.69 V. The reaction will not occur as E0cell is negative value as for spontaneous reaction value should be positive.Option C is correct.

Given below are the half-cell reactions:
$M n^{2 +} + 2 e^{-} \to M n; E^{0} = - 1.18 V$
$2 (M n^{3 +} + e^{-} \to M n^{2 +}); E^{0} = + 1.51 V$ .

The $E^{0}$ for $3 M n^{2 +} \to M n + 2 M n^{3 +}$ will be: