For a particular redox reaction MnO2 is oxidized to MnO4- and Ag+ is reduced to

Question

For a particular redox reaction MnO2 is oxidized to MnO4- and Ag+ is reduced to Ag. Complete and balance the equation for this reaction in basic solution. Phases are optional.

Explanation / Answer

To balance this equation: Divide the equation into two half-reactions: MnO2 --> MnO4- Ag+ --> Ag Since the main atoms are already balanced, balance O by adding H2O where needed: MnO2 + 2 H2O --> MnO4- Ag+ --> Ag Balance H by adding H+ MnO2 + 2 H2O --> MnO4- + 4 H+ Ag+ --> Ag Balance charges in each half reaction by adding e-: MnO2 + 2 H2O --> MnO4- + 4 H+ + 3 e- Ag+ + e- --> Ag Balance the number of electrons by multiplying 1 or both half reactions by small numbers: MnO2 + 2 H2O --> MnO4- + 4 H+ + 3 e- 3 Ag+ + 3 e- --> 3 Ag Add the two half rections: MnO2 + 3 Ag+ + 2 H2O --> 3 Ag + MnO4- + 4 H+ Now, since this reaction is occurring in a basic solution, add as many OH- as you have H+. They will form water on the right side of the equation. Subtract the waters on the left from both sides and you're done: MnO2 + 3 Ag+ + 4 OH- --> 3 Ag + MnO4- + 2 H2O

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