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 3 Mn\(^{2+}\) \(\rightarrow\) Mn + 2Mn\(^{3+}\) will be \rule{2cm{0.1mm}}

• 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:

To calculate E\(^\circ\)\(_{cell}\), you can use either of these two equivalent formulas: 1. \( E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \) (where both potentials are reduction potentials). 2. \( E^\circ_{cell} = E^\circ_{reduction} + E^\circ_{oxidation} \) (where you reverse the sign of the potential for the oxidation half-reaction). The second method is often less prone to errors. Remember that a negative E\(^\circ\)\(_{cell}\) means the reaction is not spontaneous in the forward direction.

Answer 1

The Correct Option is A

Step 1: Identify the half-reactions as they occur in the overall reaction.
The overall reaction is: 3 Mn\(^{2+}\) \(\rightarrow\) Mn + 2Mn\(^{3+}\). Let's break this down into oxidation and reduction half-reactions:

• Reduction: One Mn\(^{2+}\) ion is gaining 2 electrons to become Mn. \[ \text{Mn}^{2+} + 2e^- \rightarrow \text{Mn} \quad E^\circ_{red} = -1.18 \text{ V} \]
• Oxidation: Two Mn\(^{2+}\) ions are losing electrons to become two Mn\(^{3+}\) ions. \[ 2\text{Mn}^{2+} \rightarrow 2\text{Mn}^{3+} + 2e^- \] This is the reverse of the second given half-reaction. When we reverse a half-reaction, we change the sign of its E\(^\circ\). The given reduction is: Mn\(^{3+}\) + e\(^-\) \(\rightarrow\) Mn\(^{2+}\) (E\(^\circ\) = +1.51 V). So, the oxidation half-reaction is: Mn\(^{2+}\) \(\rightarrow\) Mn\(^{3+}\) + e\(^-\) (E\(^\circ\)\(_{ox}\) = -1.51 V).

Step 2: Calculate the standard cell potential (E\(^\circ\)\(_{cell}\)).
The standard cell potential is the sum of the standard reduction potential and the standard oxidation potential. \[ E^\circ_{cell} = E^\circ_{reduction} + E^\circ_{oxidation} \] Note: E\(^\circ\) is an intensive property, so we do not multiply it by the stoichiometric coefficients. We just add the potentials for the identified half-reactions. \[ E^\circ_{cell} = (-1.18 \text{ V}) + (-1.51 \text{ V}) = -2.69 \text{ V} \] Step 3: Determine the spontaneity of the reaction.
The spontaneity of an electrochemical reaction is determined by the sign of E\(^\circ\)\(_{cell}\).

• If E\(^\circ\)\(_{cell}\)>0, the reaction is spontaneous under standard conditions (\(\Delta\)G\(^\circ\)<0).
• If E\(^\circ\)\(_{cell}\)<0, the reaction is non-spontaneous under standard conditions (\(\Delta\)G\(^\circ\)>0).

Since E\(^\circ\)\(_{cell}\) = -2.69 V, which is negative, the reaction is non-spontaneous and will not occur as written.
Step 4: Final Answer.
The E\(^\circ\)\(_{cell}\) is -2.69 V, and the reaction is non-spontaneous.