What is the reduction half-reaction for the reaction 2MnO4- + 16H+ + 10e- → 2Mn2
Practice Questions
Q1
What is the reduction half-reaction for the reaction 2MnO4- + 16H+ + 10e- → 2Mn2+ + 8H2O?
MnO4- + 8H2O + 10e- → Mn2+ + 16H+
MnO4- + 10e- + 8H+ → Mn2+ + 4H2O
MnO4- + 10e- → Mn2+ + 8H2O
MnO4- + 16H+ → Mn2+ + 10e- + 8H2O
Questions & Step-by-Step Solutions
What is the reduction half-reaction for the reaction 2MnO4- + 16H+ + 10e- → 2Mn2+ + 8H2O?
Correct Answer: MnO4- + 8H+ + 5e- → Mn2+ + 4H2O
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.
