In acidic solution MnO4- oxidizes H3AsO3, a weak acid, to H3AsO4, a weak acid, a

Question

In acidic solution MnO4- oxidizes H3AsO3, a weak acid, to H3AsO4, a weak acid, and is reduced to Mn2+. Write the balanced net ionic equation for this reaction. How many H+ are there in the balanced equation?

Explanation / Answer

MnO4(-) + H3AsO3 --> H3AsO4 + Mn(2+) Divide the above reaction into half reactions: MnO4(-) --> Mn(2+) ... half-rxn 1 H3AsO3 --> H3AsO4 ... half-rxn 2 Now, we must balance both half reactions: MnO4(-) + 8H(+) + 5e- --> Mn(2+) + 4H2O H3AsO3 + H2O --> H3AsO4 + 2H(+) + 2e- Now, we must balance the electrons for the above equations: 2(MnO4(-) + 8H(+) + 5e- --> Mn(2+) + 4H2O) BECOMES 2MnO4(-) + 16H(+) + 10e- --> 2Mn(2+) + 8H2O 5(H3AsO3 + H2O --> H3AsO4 + 2H(+) + 2e-) BECOMES 5H3AsO3 + 5H2O --> 5H3AsO4 + 10H(+) + 10e- Balanced Net Ionic Equation: 2MnO4(-) + 16H(+) + 10e- + 5H3AsO3 + 5H2O --> 2Mn(2+) + 8H2O + 5H3AsO4 + 10H(+) + 10e- 2MnO4(-) + 6H(+) + 5H3AsO3 --> 2Mn(2+) + 3H2O + 5H3AsO4 --> This is the completely balanced net ionic equation 2 Mn on left = 2 Mn on right 7 O on left = 7 O on right 21 H on left = 21 H on right 5 As on left = 5 As on right There are 6H+ in the balanced net ionic equation. Hope this helps! :)

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