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

You visited us 1 times! Enjoying our articles? Unlock Full Access!

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 occur

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- 2.69 V

; the reaction will not occur

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

+ 2.69 V

; the reaction will occur

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- 0.33 V

; the reaction will not occur

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $- 2.69 V$ ; the reaction will not occur
Standard potential of reaction $[E^{0}]$ is given by,
$E_{0}^{c e l l} = E_{R} - E_{P}$
$M n^{2 +} + 2 e^{-} ⟶ M n E^{0} = - 1.18 V 2 M n^{2 +} ⟶ 2 M n^{3 +} + 2 e^{-} E^{0} = - 1.51 V - ------------------------------------------- -$
$3 M n^{2 +} ⟶ M n + 2 M n^{3 +}$ $E^{0} = - 1.18 + (- 1.51) = - 2.69 V$
The reaction is non spontaneous, $E^{0} = - 2.69 V$

Suggest Corrections

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 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 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. 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:

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:

Join BYJU'S Learning Program

Given below are the half-cell reactions:Mn2++2e−→Mn;Eo=−1.18V2(Mn+3+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: