What is the reduction half-reaction for the conversion of MnO4- to Mn2+ in acidi

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Question: What is the reduction half-reaction for the conversion of MnO4- to Mn2+ in acidic medium?

Options:

Correct Answer: MnO4- + 8H+ + 5e- → Mn2+ + 4H2O

Solution:

The correct reduction half-reaction in acidic medium is MnO4- + 8H+ + 5e- → Mn2+ + 4H2O.

What is the reduction half-reaction for the conversion of MnO4- to Mn2+ in acidic medium?

What is the reduction half-reaction for the conversion of MnO4- to Mn2+ in acidic medium?

Step 1: Identify the species being reduced. In this case, it is MnO4- (permanganate ion).
Step 2: Determine the oxidation state of manganese in MnO4-. It is +7.
Step 3: Identify the final product, which is Mn2+. The oxidation state of manganese in Mn2+ is +2.
Step 4: Calculate the change in oxidation state. The change is from +7 to +2, which is a reduction of 5.
Step 5: Write the half-reaction for the reduction. Start with MnO4- and add electrons to balance the charge.
Step 6: Since we are in acidic medium, add H+ ions to balance the hydrogen atoms. We need 8 H+ ions.
Step 7: Add water (H2O) to balance the oxygen atoms. We will produce 4 H2O molecules.
Step 8: Combine all parts to form the complete reduction half-reaction: MnO4- + 8H+ + 5e- → Mn2+ + 4H2O.

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