Question

For the following reaction, (1) assign oxidation numbers for all the elements, (2) determine which one is oxidized and which one is reduced, and (3) balance the redox reaction in acidic solution.

SO₃^2–(aq) + MnO₄^–(aq) ® SO₄^2–(aq) + Mn²⁺(aq)

Answer 1

Answer:-
PART(A):-
Mn+7O-24-+S+4O-232-→Mn+4O-22+S+6O-242-
PART(B):-
Reduction: MnO4- ---> Mn2+Oxidation: SO3[2-] ---> SO4[2-]In balancing redox reactions, you have to make sure that both the number of atoms and total charge on both sides of the equation are equal. Balance the reduction half first.In the reduction half, the number of Mn atoms is the same. The number of O atoms is not. There are 4 on the left andnone on the right. Use H2O molecules to balance the O atoms. In this case, 4 is needed. Put them on the right side.MnO4- ---> Mn2+ + 4H2OTo balance the H atoms, use H+ ions. Since there are 8 H atoms on the right, put 8 H+ ions on the left to make the number equal.MnO4- + 8H+ ---> Mn2+ + 4H2OThe total number of atoms on both sides is now equal. However, the total charge on both sides is not. The total charge on the left is +7, on the right, +2. To make the charges equal, place 5 electrons on the left side.MnO4- + 8H+ + 5e- ---> Mn2+ + 4H2OIn the oxidation half, do the same procedure. Use H2O to balance O atoms and H+ to balance H atoms. Use electronsto balance charge.SO3[2-] + H2O ---> SO4[2-] + 2H+ + 2e-Now put the two half-reactions together. However, you have to multiply both by a number that will equalize the number of electrons on both equations. This ensures that electrons lost is equalto electrons gained.2[MnO4- + 8H+ + 5e- ---> Mn2+ + 4H2O]5[SO3[2-] + H2O ---> SO4[2-] + 2H+ + 2e-]Use the distributive property of multiplication to multiply the number to all coefficients. You then get2MnO4 + 16H+ + 10e- ---> 2Mn2+ + 8H2O5SO3[2-] + 5H2O ---> 5SO4[2-] + 10H+ + 10e-Canceling out those stuff that appear on both sides, you now get the full equation5SO3[2-] + 2MnO4- + 6H+ ---> 5SO4[2-]+ 2Mn2+ + 3H2O
PARTC
SO2 ---> (SO4)2-MnO4- ---> (Mn)2+
You don't need to balance for S or for Mnso start with oxygen on each side. Then balance for hydrogen on each equation. Finally, put both together so your total charges cancel out (system of equationssort of).After that it's just simplification.SO2 + 2H2O ---> SO4(2-) + 4H+ +2e- ] Multiply by factor of 5MnO4- + 8H+ +5e----> Mn(2+) + 4H2O ] Multiply by factor of 25SO2 + 10H2O ----> 5SO4 + 20H + 10e-2MnO4 + 16H + 10e- ----> 2Mn(2+)+8H2Oelectrons cancel out, now just simplify it all out.5SO2 + 2MnO4(-) + 2H2O ----> 5SO4(-) + 2Mn(2+) + 4H2O