Question

Given below are the half-cell reactions:
$Mn^{2+} + 2e^- \rightarrow Mn$ (E$^\circ$ = -1.18 V)
$Mn^{3+} + e^- \rightarrow Mn^{2+}$ (E$^\circ$ = +1.51 V)
The E$^\circ_{cell}$ for $3Mn^{2+} \rightarrow Mn + 2Mn^{3+}$ will be \textunderscore\textunderscore\textunderscore\textunderscore

• A. - 2.69 V, the reaction will not occur (Non-Spontaneous)
• B. - 2.69 V, the reaction will occur (Spontaneous)
• C. - 0.33 V, the reaction will not occur (Non-Spontaneous)
• D. - 0.33 V, the reaction will occur (Spontaneous)

Hint:

Always ensure consistency when adding potentials: either sum reduction potential of cathode and oxidation potential of anode, or subtract reduction potential of anode from reduction potential of cathode. For spontaneity, a positive $E^\circ_{cell}$ (or negative $\Delta G^\circ$) is required.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks to calculate the standard cell potential (E$^\circ_{cell}$) for a given redox reaction involving manganese ions and determine its spontaneity. We are provided with the standard reduction potentials for two relevant half-reactions.
Step 2: Key Formula or Approach:
1. Identify the oxidation and reduction half-reactions in the target overall reaction.
2. Assign the standard reduction potentials to the appropriate cathode and anode half-reactions. If a given half-reaction needs to be reversed for the target reaction, its potential's sign must be flipped (though it's often safer to use $\Delta G^\circ$ for non-standard combinations or to consistently assign $E^\circ_{red}$ for cathode and $E^\circ_{ox}$ for anode).
3. Calculate $E^\circ_{cell} = E^\circ_{\text{cathode}} + E^\circ_{\text{anode}}$ (where $E^\circ_{\text{anode}}$ is the standard oxidation potential). Or, $E^\circ_{cell} = E^\circ_{\text{red,cathode}} - E^\circ_{\text{red,anode}}$.
4. Determine spontaneity: If $E^\circ_{cell}>0$, the reaction is spontaneous. If $E^\circ_{cell}<0$, it is non-spontaneous.
Step 3: Detailed Explanation:
Target reaction: $3Mn^{2+} \rightarrow Mn + 2Mn^{3+}$
This is a disproportionation reaction where $Mn^{2+}$ is simultaneously reduced to $Mn$ and oxidized to $Mn^{3+}$.
Let's break down the overall reaction into half-reactions:
1. Reduction part: $Mn^{2+} + 2e^- \rightarrow Mn$
The standard reduction potential for this is given as $E^\circ_{\text{red,1}} = -1.18 \text{ V}$.
2. Oxidation part: $Mn^{2+} \rightarrow Mn^{3+} + e^-$
The given reaction is $Mn^{3+} + e^- \rightarrow Mn^{2+}$, with $E^\circ_{\text{red,2}} = +1.51 \text{ V}$.
To get the oxidation $Mn^{2+} \rightarrow Mn^{3+} + e^-$, we must reverse the second given half-reaction. The standard oxidation potential for this reaction will be the negative of its standard reduction potential: $E^\circ_{\text{ox}} = -E^\circ_{\text{red,2}} = -1.51 \text{ V}$.
Now, combine these two half-reactions to match the overall stoichiometry. The overall reaction needs 2 moles of $Mn^{3+}$ formed, so the oxidation half-reaction must occur twice, leading to a transfer of 2 electrons. The reduction half-reaction involves 2 electrons.
So, the balanced half-reactions are:
Cathode (Reduction): $Mn^{2+} + 2e^- \rightarrow Mn \quad (E^\circ_{\text{red}} = -1.18 \text{ V})$
Anode (Oxidation): $2Mn^{2+} \rightarrow 2Mn^{3+} + 2e^- \quad (E^\circ_{\text{ox}} = -1.51 \text{ V})$
(Note: $E^\circ$ values are intensive properties and do not change when the stoichiometric coefficients are multiplied.)
Now, calculate $E^\circ_{cell}$:
\[ E^\circ_{cell} = E^\circ_{\text{red}} + E^\circ_{\text{ox}} \]
\[ E^\circ_{cell} = (-1.18 \text{ V}) + (-1.51 \text{ V}) \]
\[ E^\circ_{cell} = -2.69 \text{ V} \]
Spontaneity:
Since $E^\circ_{cell}$ is negative ($-2.69 \text{ V}$), the Gibbs free energy change ($\Delta G^\circ = -nFE^\circ_{cell}$) will be positive. A positive $\Delta G^\circ$ indicates a non-spontaneous reaction under standard conditions.
Step 4: Final Answer:
The E$^\circ_{cell}$ for the reaction is -2.69 V, and the reaction will not occur (Non-Spontaneous). This matches option (1).