Balance the following equation in basic solution. Indicate what is oxidized, wha

Question

Balance the following equation in basic solution. Indicate what is oxidized, what is reduced, what is the oxidizing agent, and what is the reducing agent. MnO4- + CN- ? MnO2 + CNO-

Explanation / Answer

First, split the reaction into half reactions: MnO4(-) --> MnO2 CN(-) --> CNO(-) Now, we balance both half reactions above by adding electrons, H2O, and H(+) ions: MnO4(-) + 4H(+) + 3e- --> MnO2 + 2H2O ...MnO4(-) is reduced because the electrons are on the left side and it is the oxidizing agent. CN(-) + H2O --> CNO(-) + 2H(+) + 2e- ...CN(-) is oxidized because the electrons are on the right side and it is the reducing agent. Now, we multiply equation 1 by 2 and equation 2 by 3 to make 6e- in both equations: 2MnO4(-) + 8H(+) + 6e- --> 2MnO2 + 4H2O 3CN(-) + 3H2O --> 3CNO(-) + 6H(+) + 6e- Complete equation: 2MnO4(-) + 8H(+) + 3CN(-) + 3H2O + 6e- --> 2MnO2 + 4H2O + 3CNO(-) + 6H(+) + 6e- 2MnO4(-) + 2H(+) 3CN(-) --> 2MnO2 + H2O + 3CNO(-) ...in a basic solution, add OH(-) to both sides of the equation 2MnO4(-) + 2H(+) 3CN(-) + 2OH(-) --> 2MnO2 + H2O + 3CNO(-) + 2OH(-) 2MnO4(-) + 3CN(-) + 2H2O --> 2MnO2 + H2O + 3CNO(-) + 2OH(-) ...2H(+) + 2OH(-) form 2 H2O molecules 2MnO4(-) + 3CN(-) + 2H2O --> 2MnO2 + H2O + 3CNO(-) + 2OH(-) 2MnO4(-) + 3CN(-) + H2O --> 2MnO2 + 3CNO(-) + 2OH(-) Hope this helps! :)

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