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

To determine the E\(^\circ\)\(_{cell}\) for the reaction: 3 Mn\(^{2+}\) \(\rightarrow\) Mn + 2Mn\(^{3+}\), we begin by analyzing the given half-cell reactions and their standard electrode potentials:

• Mn\(^{2+}\) + 2e\(^-\) \(\rightarrow\) Mn (E\(^\circ\) = -1.18 V)
• Mn\(^{3+}\) + e\(^-\) \(\rightarrow\) Mn\(^{2+}\) (E\(^\circ\) = +1.51 V)

Now, rewrite the overall reaction based on stoichiometry:

The overall reaction is 3 Mn\(^{2+}\) \(\rightarrow\) Mn + 2Mn\(^{3+}\).

This involves two processes:

1. Reduction process: Mn\(^{2+}\) + 2e\(^-\) \(\rightarrow\) Mn
2. Oxidation process: Mn\(^{2+}\) \(\rightarrow\) Mn\(^{3+}\) + e\(^-\)

To balance the electrons in the oxidation and reduction steps, consider:

1. 2 electrons are needed in step 1, so multiply step 2 by 2.

Hence, the oxidation half-reaction becomes:

2 Mn\(^{2+}\) \(\rightarrow\) 2 Mn\(^{3+}\) + 2e\(^-\) (Adjusted for stoichiometry)

The combined reaction becomes:

Mn\(^{2+}\) + 2 Mn\(^{2+}\) \(\rightarrow\) Mn + 2 Mn\(^{3+}\).

Now calculate the overall E\(^\circ\)\(_{cell}\):

The net reaction is derived from the given half-cell reactions:

E\(^\circ\)\(_{cell}\) = E\(^\circ\)\(_{cathode}\) - E\(^\circ\)\(_{anode}\) = (-1.18 V) - (1.51 V)

= -1.18 V - 1.51 V = -2.69 V

The negative value indicates that the reaction is non-spontaneous under standard conditions.

Thus, the correct answer is: - 2.69 V, the reaction will not occur (Non-Spontaneous).