write the total and net ionic equations for 2KMnO4 + 16 HCl--->2MNCl2 +2 KCl +5C

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Explanation / Answer

There is no Oxygen on the RHS of your equation, so there is an error here. I suspect that your H2 should be H2O KMnO4(aq) + HCl(aq) -----> KCl(aq) + MnCl2(aq) + H2O(l) + Cl2(g) I would then write out the TOTAL IONIC equation, so that you can see what is going on with the MnO4- ion, what is it being reduced to? K+(aq) + MnO4-(aq) + H+(aq) + Cl-(aq) -----> K+(aq) + Cl-(aq) + Mn2+(aq) 2Cl-(aq) + H2O(l) + Cl2(g) You can now see that MnO4- is being converted to Mn2+ In most chem books you will find a list of Standard electrode potentials, this gives you a list of common reduction half equations, MnO4- can be reduced in several ways, pick the half equation that fits your equation, this will be MnO4-(aq) + 8H+(aq) + 5e -----> Mn2+(aq) + 4H2O(l) This is gain of electrons (e is on the LHS of the equation) Therefore this is the reduction half equation (reduction is gain of electrons. The species that is losing electrons (oxidation) is Cl- 2Cl- -----> Cl2(g) + 2e This is your oxidation half equation Add them together.... You will notice that there are 5 e in your reduction and only 2 in you oxidation half equation... This needs to be balanced in the same way you would balance a normal equation, cross multiply 2 x ( MnO4-(aq) + 8H+(aq) + 5e -----> Mn2+(aq) + 4H2O(l)) So , balanced reduction step = 2MnO4-(aq) + 16H+(aq) + 10e ---> 2Mn2+(aq) + 8H2O 5 x (2Cl- -----> Cl2(g) + 2e) Balanced oxidion step = 10Cl-(aq) -----> 5Cl2(g) + 10 e Now there are 10 e for both the oxidation and reduction steps of the equation, so they are balanced. They cancel out. Now add the two together This is the balanced Net Ionic equation 2MnO4-(aq) + 16H+(aq) + 10Cl-(aq) ----> 2Mn2+(aq) + 8H2O(l) + 5Cl2(g) The NET Ionic equation does not take into account the spectator ions. Spectator ions are those that do not take part in the reaction at all. This equation is a bit tricky because some of the Cl- ions do take part in the reaction and some do not. The Cl- ions in both the KCl and MnCl2 are spectator ions. This is why you have 16H+(aq) and only 10 Cl-(aq) in the NET equation, which might confuse you because everyone knows that HCl has one atom of each. However, when you write out the full molecular equation including the K in the KMnO4, KCl(aq) and the Cl in the MnCl2(aq) stick to the same numbers as you found for the NET ionic equation and it will all work out. 2KMnO4(aq) + 16HCl(aq) ----> 2KCl(aq) + 2MnCl2(aq) + 8H2O(l) + 5Cl2(g) There are 16 Cl on both sides There are 16 H on both sides etc So it is balanced OK, to summarise Balanced reduction half equation 2MnO4-(aq) + 16H+(aq) + 10e ---> 2Mn2+(aq) + 8H2O Balanced oxidation half equation 10Cl-(aq) -----> 5Cl2(g) + 10 e Balanced NET IONIC equation 2MnO4-(aq) + 16H+(aq) + 10Cl-(aq) ----> 2Mn2+(aq) + 8H2O(l) + 5Cl2(g) Balanced molecular equation 2KMnO4(aq) + 16HCl(aq) ----> 2KCl(aq) + 2MnCl2(aq) + 8H2O(l) + 5Cl2(g)

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