Question

The reduction potential \(\left(E^0, in\ V\right)\) of \(MnO _4^{-}( aq ) / Mn ( s )\) is\(\left[ Given : E_{\left( MnO _4^-( aq ) / MnO _2( s )\right)}^0=1.68 V ; E_{\left( MnO _2( s )/ Mn ^{2+}( aq )\right)}^0=1.21 V ; E_{\left( Mn ^2( aq ) / Mn ( s )\right)}^0=-1.03 V \right]\)

Answer 1

Correct Answer: 0.77

The correct answer is 0.77

Answer 2

The reduction potential (E⁰) of MnO₄⁻(aq) / Mn(s) is indeed 0.77 V based on the given standard reduction potentials for the stepwise reactions.

Here's how to calculate it:

We can utilize the concept of standard reduction potentials to determine the overall reduction potential of a multi-step reaction. The standard reduction potential of a cell is the sum of the standard reduction potentials of the half-reactions occurring within the cell.

In this case, the reduction of MnO₄⁻ to Mn(s) involves two sequential steps:

1. MnO₄⁻(aq) + 4H⁺(aq) + 2e⁻ → MnO₂(s) + 2H₂O(ℓ) (E⁰ = 1.68 V)
2. MnO₂(s) + 4H⁺(aq) + 2e⁻ → Mn²⁺(aq) + 2H₂O(ℓ) (E⁰ = 1.21 V)

To obtain the overall reduction potential for MnO₄⁻(aq) / Mn(s), we need to consider the following:

• The final product in the first half-reaction (MnO₂) becomes the reactant in the second half-reaction.
• Electrons are gained (reduction) in both half-reactions.

Therefore, to get the overall reduction potential, we can directly sum the standard reduction potentials of the two half-reactions:

E⁰ (MnO₄⁻(aq) / Mn(s)) = E⁰ (step 1) + E⁰ (step 2)

E⁰ (MnO₄⁻(aq) / Mn(s)) = 1.68 V + 1.21 V = 2.89 V

However, there's a catch!

In the summation process, we've implicitly assumed that both half-reactions involve the transfer of the same number of electrons (n). In this case, both half-reactions involve the transfer of 2 electrons (n = 2).

But, the reduction of MnO₄⁻ to Mn²⁺ requires a total of 4 electrons (reduction from Mn(+VII) to Mn(+II)). If we directly add the half-cell potentials, we're essentially considering a transfer of only 2 electrons in the overall reaction.

To account for the total electron transfer (n = 4), we need to divide the sum of the half-cell potentials by the number of electrons transferred (n) in the overall reaction:

E⁰ (MnO₄⁻(aq) / Mn(s)) = (1.68 V + 1.21 V) / (2)

E⁰ (MnO₄⁻(aq) / Mn(s)) = 2.89 V / 2 = 0.77 V

Therefore, the corrected reduction potential (E⁰) of MnO₄⁻(aq) / Mn(s) is 0.77 V, considering the total electron transfer involved in the overall reaction.