Question

Reaction between potassium permanganate and sodium thiosulfate

Answer 1

ANSWER=

Basically the reaction equation is be correct. In strong acidic solution permanganate is reduced to manganese ions and thiosulfate becomes oxidized to sulfate.

KMnO₄ + H₂SO₄ + Na₂S₂O₃ → K₂SO₄ + MnSO₄ + H₂O + 5 Na₂SO₄

You have 10 sodium atoms comprised in sodium sulfate on right hand side. The only reactant containing sodium is Na₂S₂O₃. For 10 Na atoms it must have stoichiometric factor 5:

KMnO₄ + H₂SO₄ + 5 Na₂S₂O₃ → K₂SO₄ + MnSO₄ + H₂O + 5 Na₂SO₄

Next step to balance these equation for a redox reaction is balance the number of electrons transferred from the oxidized species to the reduced species.

The two sulfur atoms in thiosulfate have oxidation number + II and one sulfate atom in sulfate has +VI. So each sulfur atom gives away 4 electrons

S₂O₃²⁻→ 2 SO₄²⁻+ 8 e⁻

Note This half reaction is unbalanced in O, i.e. 5 O atoms ar missing on left hand side. Don't worry about it. This is fixed later.

We consider the oxidation of five thiosulfate ions. so the oxidation half reaction is:

5 S₂O₃²⁻→ 10 SO₄²⁻+ 40 e⁻

So we need 40 electrons from reduction half reaction. Manganese in permanganate has oxidation number + VII and as simple ion it has + II. Hence,

MnO₄⁻ + 5 e⁻ → Mn²⁺

In order to get the 40 electrons you have to multiply be 8:

8 MnO₄⁻ + 40 e⁻ → 8 Mn²⁺

So the stoichiometric coefficient for potassium permanganate is 8

8 KMnO₄ + H₂SO₄ + 5 Na₂S₂O₃ → K₂SO₄ + MnSO₄ + H₂O + 5 Na₂SO₄

For both potassium and manganese 8 atoms are found on left hand side. Thes two elements are only found in their respective sulfates on left hand side. To get same number of K and Mn on left hand side, their stoichiometric factors must be 4 and 8 respectively.

8 KMnO₄ + H₂SO₄ + 5 Na₂S₂O₃ → 4 K₂SO₄ + 8 MnSO₄ + H₂O + 5 Na₂SO₄

Then count the sulfur atoms. On right hand side there are

4 + 8 + 5 = 17

On left hand side there are the 10 from thiosulfate. So the must be 7 sulfur atoms from the sulfuric acid, hence

8 KMnO₄ + 7 H₂SO₄ + 5 Na₂S₂O₃ → 4 K₂SO₄ + 8 MnSO₄ + H₂O + 5 Na₂SO₄

In right to get identical number of H atoms water must have same stoichiometric factor as sulfuric ad so

8 KMnO₄ + 7 H₂SO₄ + 5 Na₂S₂O₃ → 4 K₂SO₄ + 8 MnSO₄ + 7 H₂O + 5 Na₂SO₄

Now let's check whether the number oxygen atoms is in balance or not.

number of O on LHS

8∙4 + 7∙4 + 5∙3 = 75

number of O on RHS

4∙4 + 8∙4 + 7∙1 + 5∙ 4 = 75

perfect. ;-)

So the balanced reaction equation is:

8 KMnO₄ + 7 H₂SO₄ + 5 Na₂S₂O₃ → 4 K₂SO₄ + 8 MnSO₄ + 7 H₂O + 5 Na₂SO₄