Question

Write down the balanced net ionic equation

Permanganate ion reacts with thiosulfate ion in acidic condition to produce Mn2+ ion and sulfate ion.

Answer 1

Write the two half reactions,

1) first reduction half reaction

MnO4- -------> Mn2+

2) Mn is going from +7 to +2 oxidation state by gain of 5e-, balance charges

MnO4- + 5e- -------> Mn2+

3) Balance oxygen by adding water to the other side

MnO4- + 5e- -------> Mn2+ + 4H2O

4) balance hydrogen by adding H+ on the other side (reaction in acidic medium)

MnO4- + 8H+ + 5e- -------> Mn2+ + 4H2O

5) write the other half rection

S2O3(2-) ---------> SO4(2-)

6) balance sulfur on both sides

S2O3(2-) ---------> 2SO4(2-)

7) sulfur is going from +2 to +6 and looses 4e-, balance charge

S2O3(2-) ---------> 2SO4(2-) + 4e-

8) balance oxygen by adding water to other side

S2O3(2-) + 4H2O ---------> 2SO4(2-) + 3H2O + 4e-

9) balance hydrogen by adding H+ on the other side

S2O3(2-) + 4H2O + 6H+ ---------> 2SO4(2-) + 3H2O + 8H+ + 4e-

thus, net half reaction will be,

S2O3(2-) + H2O ---------> 2SO4(2-) + 2H+ + 4e-

10) balance charges on the two half reaction,

(MnO4- + 8H+ + 5e- -------> Mn2+ + 4H2O) x 4

(S2O3(2-) + H2O ---------> 2SO4(2-) + 2H+ + 4e-) x 5

11) net reaction will be

4MnO4- + 32H+ + 20e- -------> 4Mn2+ + 16H2O

5S2O3(2-) + 5H2O ---------> 10SO4(2-) + 10H+ + 20e-

12) add the two half reaction and cancel the common terms we get the ne ionic reaction,

4MnO4-(aq) + 5S2O3(2-)(aq) + 22H+(aq) --------> 4Mn2+(s) + 10SO4(2-)(aq) + 11H2O(l)