Chemistry

Chemistry

Problem

The two half-reactions below occur in acidic solution. Oxidation: Sn2+→Sn4++2e−\mathrm{Sn^{2+} \rightarrow Sn^{4+} + 2e^-} Reduction: MnO4−+8H++5e−→Mn2++4H2O\mathrm{MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O} Which overall equation is correctly balanced for both atoms and net charge? a) 5Sn2++2MnO4−+8H+→5Sn4++2Mn2++4H2O\mathrm{5Sn^{2+} + 2MnO_4^- + 8H^+ \rightarrow 5Sn^{4+} + 2Mn^{2+} + 4H_2O} b) Sn2++MnO4−+8H+→Sn4++Mn2++4H2O\mathrm{Sn^{2+} + MnO_4^- + 8H^+ \rightarrow Sn^{4+} + Mn^{2+} + 4H_2O} c) 5Sn2++2MnO4−+16H+→5Sn4++2Mn2++8H2O\mathrm{5Sn^{2+} + 2MnO_4^- + 16H^+ \rightarrow 5Sn^{4+} + 2Mn^{2+} + 8H_2O} d) Sn2++2MnO4−+16H+→Sn4++2Mn2++8H2O\mathrm{Sn^{2+} + 2MnO_4^- + 16H^+ \rightarrow Sn^{4+} + 2Mn^{2+} + 8H_2O}
Intuitive explanation

This is about combining half-reactions into a balanced redox equation—the key is matching electron transfer. You need to find the smallest whole-number multipliers for each half-reaction so the electrons lost in oxidation equal the electrons gained in reduction. Once the electron counts match, multiply each half-reaction accordingly, then add them together and cancel any species that appear on both sides. Check that atoms and charge balance in your final equation.

Intuitive explanation

This is about combining half-reactions into a balanced redox equation—the key is matching electron transfer. You need to find the smallest whole-number multipliers for each half-reaction so the electrons lost in oxidation equal the electrons gained in reduction. Once the electron counts match, multiply each half-reaction accordingly, then add them together and cancel any species that appear on both sides. Check that atoms and charge balance in your final equation.

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