Consider the half-cell reduction reaction :- Mn2++2e-→ Mn, E0=-1.18 V Mn2+→ Mn3++e-, E0=-1.51 V The E° for the reaction 3 Mn2+→ Mn0+2Mn3+ and possibility of the forward reaction are respectively :

Question

Consider the half-cell reduction reaction :- Mn2++2e-→ Mn, E0=-1.18 V Mn2+→ Mn3++e-, E0=-1.51 V The E° for the reaction 3 Mn2+→ Mn0+2Mn3+ 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

Tardigrade Admin · Accepted Answer

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

Δ G

will come +ve for this case and that indicates reaction is non-spontaneous.

Q. Consider the half-cell reduction reaction :- Mn2++2e−→Mn,E0=−1.18V Mn2+→Mn3++e−,E0=−1.51V The E∘ for the reaction 3Mn2+→Mn0+2Mn3+ and possibility of the forward reaction are respectively :

- 4.18 V and yes

+ 0.33 V and yes

+ 2.69 V and no

- 2.69 V and no

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