#1. What is the coefficient of Fe2+ when the following redox equation is balance

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Oxidation is loss of electrons Reduction is gain of electrons The tricky part of this one is knowing how to balance the MnO4- half equation MnO4- -------------> Mn2+ You can see there is O in MnO4- but none on the other side of the arrow. To balance it you need to add H2O to the right hand side of the arrow MnO4- -------------> Mn2+ + H2O But now you need to add H+ to the other side, to counter the H in the H2O you added MnO4- + H+ -----> Mn2+ + H2O Now, balance it MnO4- + 8 H+ ------> Mn2+ + 4 H2O To work out how many electrons are in the equation we need to balance the charges on each side of the arrow There is 1 x MnO4- and 8 x H+ on the left = 7+ There is 1 x Mn2+ on the right side = 2+ So we need to add 7 - 2 = 5 electrons on the right hand side Reduction half equation is therefore::: MnO4-(aq) + 8H+(aq) + 5 e -----> Mn2+(aq) + 4H2O(l) The Fe is much easier Oxidation half equation Fe2+ ---------------> Fe3+ + 1 e Now we need to balance the electrons between the oxidation half equation and the reduction half equation. There are 5 electrons in the reduction half equation and only one in the oxidation half equation. So we need to multiply the oxidation half equation by 5, to get 5 e 5Fe2+ -------------> 5Fe3+ + 5e Now, add the two equations together MnO4- + 8H+ + 5Fe2+ + 5e ---> 5Fe3+ + Mn2+ + 4H2O + 5e The 5 e on each side will cancel out, so we can remove them MnO4-(aq) + 8H+(aq) + 5Fe2+(aq) ---> 5Fe3+(aq) + Mn2+(aq) + 4H2O(l)

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