Question

When 0.1 mol MnO₄^2– is oxidised the quantity of electricity required to completely oxidise MnO4^2– to MnO₄^– is

• A. 2 x 96500 C
• B. 9650 C
• C. 96.50 C
• D. 96500 C

Answer 1

The Correct Option is B

To determine the quantity of electricity required to completely oxidize MnO_4^{2-} to MnO_4^{-}, we need to understand the change in oxidation state and use Faraday’s Laws of Electrolysis.

The oxidation of MnO_4^{2-} to MnO_4^{-} can be represented as:

MnO_4^{2-} \rightarrow MnO_4^{-} + e^–

From the equation, it is clear that each mole of MnO_4^{2-} requires 1 mole of electrons to be converted to MnO_4^{-}. Thus, 1 electron is involved in the reaction.

According to Faraday’s laws, the quantity of electricity (Q) required to transfer a mole of electrons is given by Q = n \cdot F, where:

• n = number of moles of electrons
• F = Faraday’s constant, approximately 96500 \, C/mol

In this case, for 0.1 mol of MnO_4^{2-}, the number of moles of electrons required is:

n = 0.1 \, mol \times 1 = 0.1 \, mol

Now, calculate the quantity of electricity:

Q = 0.1 \, mol \times 96500 \, C/mol = 9650 \, C

Thus, the correct answer is 9650 \, C.

Conclusion: The amount of electricity required to oxidize 0.1 mol of MnO_4^{2-} to MnO_4^{-} is 9650 C, making the correct answer 9650 C.