What is the reduction half-reaction for the reaction 2MnO4- + 16H+ + 10e- → 2Mn2+ + 8H2O?

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Q: What is the reduction half-reaction for the reaction 2MnO4- + 16H+ + 10e- → 2Mn2+ + 8H2O?

Solution: The correct reduction half-reaction shows the gain of electrons and protons.

Steps: 5

Step 1: Identify the reactants and products in the given reaction: 2MnO4-, 16H+, and 10e- are the reactants, while 2Mn2+ and 8H2O are the products.
Step 2: Recognize that a reduction half-reaction involves the gain of electrons. In this case, 10 electrons (10e-) are being gained.
Step 3: Note that protons (H+) are also involved in the reaction, indicating that the reaction occurs in an acidic environment.
Step 4: Write the reduction half-reaction by showing the reactants on the left side and the products on the right side, including the gained electrons and protons.
Step 5: The final reduction half-reaction is: 2MnO4- + 16H+ + 10e- → 2Mn2+ + 8H2O.