Question

Consider the half-cell reduction reaction :- $Mn^{2+}+2e^{-}\to Mn, E^{0}=-1.18 V$ $Mn^{2+}\to Mn^{3+}+e^{-}, E^{0}=-1.51 V$ The $E^\circ$ for the reaction $3 Mn^{2+}\to Mn^{0}+2Mn^{3+}$ and possibility of the forward reaction are respectively :

• A. - 4.18 V and yes
• B. + 0.33 V and yes
• C. + 2.69 V and no
• D. - 2.69 V and no

Answer 1

The Correct Option is D

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 nd 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 $\Delta G$ will come +ve for this case and that indicates reaction is non-spontaneous.