S2O5^2- + MnO4^2- + Mn2O3 Solution sol S2O5 ==> SO4 2- . . .and MnO4- ==> Mn2O3

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sol S2O5 ==> SO4 2- . . .and MnO4- ==> Mn2O3 . . .balance each one separately. S2O5 ==> SO4 2- . . .put a 2 in front of SO4 to balance S. S2O5 ==> 2SO4 2- . . . put 3H2O on the left side to balance O. S2O5 + 3H2O ==> 2SO4 2-. . .put 6H+ on the right side to balance H. S2O5 + 3H2O ==> 2SO4 2- + 6H+ . . .put 2e- on the right side to balance the charge. S2O5 + 3H2O ==> 2SO4 2- + 6H+ + 2e- MnO4- ==> Mn2O3 . . .put a 2 in front of MnO4- to balance Mn. 2MnO4- ==> Mn2O3 . . .put 5H2O on the right side to balance O. 2MnO4- ==> Mn2O3 + 5H2O . . .put 10H+ on the left side to balance H. 2MnO4- + 10H+ ==> Mn2O3 + 5H2O . . .put 8e- on the left side to balance the charge. 2MnO4- + 10H+ + 8e- ==> Mn2O3 + 5H2O the first reaction has 2e- on the right side while the last one has 8e- on the left side. Multiply the first reaction by 4 to give 8e- on the right side, so that when we add the two equations together, the e- will cancel out. . 4(S2O5 + 3H2O ==> 2SO4 2- + 6H+ + 2e-) + 2MnO4- + 10H+ + 8e- ==> Mn2O3 + 5H2O ================================== = 4S2O5 + 12H2O + 2MnO4- + 10H+ ==> 8SO4 2- + 24H+ + Mn2O3 + 5H2O Cancel 5H2O and 10H+ from both sides. 4S2O5 + 7H2O + 2MnO4- ==> 8SO4 2- + 14H+ + Mn2O3

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