What is the reduction half-reaction for the reaction of MnO4- in acidic medium?
Practice Questions
Q1
What is the reduction half-reaction for the reaction of MnO4- in acidic medium?
MnO4- + 8H+ + 5e- → Mn2+ + 4H2O
MnO4- + 3e- → MnO2 + 2H2O
MnO4- + 2e- → MnO2 + 4H+
MnO4- + 4e- + 8H+ → MnO2 + 4H2O
Questions & Step-by-Step Solutions
What is the reduction half-reaction for the reaction of MnO4- 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 product of the reduction. In acidic medium, MnO4- is reduced to Mn2+.
Step 4: Write the half-reaction for the reduction. Start with MnO4- and show it gaining electrons to become Mn2+.
Step 5: Balance the charge by adding electrons. Since MnO4- has a charge of -1 and Mn2+ has a charge of +2, you need to add 8 electrons to the left side: MnO4- + 8e- → Mn2+.
Step 6: Balance the oxygen atoms by adding water molecules. There are 4 oxygen atoms in MnO4-, so add 4 H2O to the right side: MnO4- + 8e- → Mn2+ + 4H2O.
Step 7: Balance the hydrogen atoms by adding H+ ions. There are 8 hydrogen atoms in 4 H2O, so add 8 H+ to the left side: MnO4- + 8e- + 8H+ → Mn2+ + 4H2O.
