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

To solve this problem, we need to determine the standard cell potential \( E^\circ \) for the reaction: 3 Mn^{2+} \to Mn^{0} + 2 Mn^{3+}.

For this, let's break down the problem step-by-step:

1. The provided half-cell reactions are:
   • Mn^{2+} + 2e^{-} \to Mn,\quad E^\circ = -1.18\,V
   • Mn^{2+} \to Mn^{3+} + e^{-},\quad E^\circ = -1.51\,V
2. We need to construct a balanced redox reaction: 3Mn^{2+} \to Mn^{0} + 2Mn^{3+}.
3. Considering the given reactions, we can rearrange:
   • From Mn^{2+} + 2e^{-} \to Mn, reverse it to get oxidation: Mn \to Mn^{2+} + 2e^{-} with E^\circ = +1.18\,V.
   • Keep Mn^{2+} \to Mn^{3+} + e^{-} as it is.
4. Now balance these reactions:
   • Multiply the Mn^{2+} \to Mn^{3+} reaction by 2: 2Mn^{2+} \to 2Mn^{3+} + 2e^{-}.
   • Add it to the reversed reaction: Mn \to Mn^{2+} + 2e^{-}.
5. Combining the reactions:
   • Overall reaction achieved: 3Mn^{2+} \to Mn + 2Mn^{3+}.
   • The net electrode potential is: E^\circ = +1.18\,V + (-1.51\,V \times 2)\,.
6. Calculate the standard electrode potential:
   • E^\circ = +1.18 - 3.02 = -2.69\,V.
7. A negative \( E^\circ \) implies the reaction is non-spontaneous in the forward direction.

Thus, the correct answer is:
E^\circ = -2.69\,V and the forward reaction is not feasible (no).