balance net ionic equation BASIC SOLUTION MnO 4 -(aq) +Fe +2 (aq)--->Fe +3 (aq)+

Question

balance net ionic equation BASIC SOLUTION MnO4-(aq) +Fe+2(aq)--->Fe+3(aq)+MnO2(s) balance net ionic equation BASIC SOLUTION MnO4-(aq) +Fe+2(aq)--->Fe+3(aq)+MnO2(s)

Explanation / Answer

Separate into oxidationand reduction half reactions Oxidation: Fe2+ -->Fe3+ Reduction:MnO4- -->MnO2 Balance allelements besides oxygen and hydrogen. Then balance oxygen by addingwater to the side that isdeficient Oxidation: Fe2+ --> Fe3+ Reduction: MnO4---> MnO2 + 2H2O Balance the H by adding H+ to the sidethat isdeficientOxidation: Fe2+ --> Fe3+ Reduction: MnO4- +4H+--> MnO2 + 2H2O Balance charge by adding electrons to themore positive side of theequations Oxidation: Fe2+--> Fe3+ + 1e- Reduction: MnO4- +4H+ + 3e- --> MnO2 +2H2O Multiply a half-reaction by an integervalue so that the number of electrons is thesameOxidation: 3[Fe2+ --> Fe3+ + 1e-] =>3Fe2+--> 3Fe3+ + 3e- Reduction: MnO4- + 4H+ + 3e---> MnO2 + 2H2O Reaction takes place underbasic conditions. Add equivalent amount of OH- to both sides ofreaction to cancel out the H+. OH- and H+form H2O when on the sameside.Oxidation:3Fe2+ -->3Fe3+ + 3e- Reduction: MnO4- +4H2O +3e- --> MnO2+ 2H2O +4OH- Combine reactions into one. Cancel whatoccurs simultaneously on the left andright 3Fe2++ MnO4- +2H2O + 3e---> 3Fe3+ + MnO2 +2H2O + 4OH- +3e-

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