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Consider the half-cell reduction reaction:

Mn^{2+}+2e^{-}\rightarrow Mn, \: E^{\circ}=-1.18V

Mn^{2+}\rightarrow Mn^{3+}+e^{-}, \: E^{\circ}=-1.51V

The E° for the reaction 3Mn^{2+}\rightarrow Mn^{0}+2Mn^{3+}, and possibility of the forward reaction are, respectively

  • Option 1)

    -2.69V and no

  • Option 2)

    -4.18 V and yes

  • Option 3)

    +0.33 V and yes

  • Option 4)

    +2.69 V and no

 

Answers (1)

 

Gibbs energy of the reaction -

\Delta_{r}G=-nFE_{cell}

- wherein

\Delta _{r}G = gibbs energy of the reaction

E_{cell}= emf of the cell

nF = amount of charge passed

 

 Mn^{2+}+2e^{-}\rightarrow Mn

E^\circ=-1.18 V

2Mn^{2+} \longrightarrow 2Mn^{3+}+2e\: E^\circ=-1.51V

For the cell

3MN^{+2} \longrightarrow Mn+ 2Mn^{3},\: E^\circ = -2.69 V

Since E^\circ is negative, so the process is non spontaneous


Option 1)

-2.69V and no

This is correct option

Option 2)

-4.18 V and yes

This is incorrect option

Option 3)

+0.33 V and yes

This is incorrect option

Option 4)

+2.69 V and no

This is incorrect option

Posted by

Vakul

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