Write the balanced equation for the overall reaction in acidic solution. Cr₂O₇²⁻(aq) + S₂O₃²⁻ (aq) → Cr³⁺ (aq) + S₄O₆²⁻ (aq) ANSWER QUICK PLEASE


Step 1 of 1

Step 1

Here given unbalanced equation ,

Cr2O72- + S2O32- ⟶ Cr3+ + S4O62-

Objective :

Balance the given redox equation with a suitable method.

Concept:

Use the oxidation number method here.

Final solution

Step-2

  1. We write the skeleton equation of all the reactants and products of the reaction.

Cr2O72- + S2O32- ⟶ Cr3+ + S4O62-

  1. Indicate the oxidation number of each element above its symbol and identify which elements which undergoes a change in the oxidation number.

+6 +2 +3 +2.5

Cr2O72- + S2O32- ⟶ Cr3+ + S4O62-

  1. Calculate the increase or decrease in oxidation number per atom (n-factor =nf) and identify the oxidizing and reducing agent.

nf =( Oxidation number of product – Oxidation number of reactant) * Number of atoms

Note – This formula is applied only when if more than one atom of same elements is present.

+6 +3

Cr2O72- ⟶ Cr3+ ( Reduction ) So Cr2O72- is oxidizing agent

nf = (3-6)*2=-6

+2 +2.5

S2O32- ⟶ S4O62- (Oxidation) So S2O32- is reducing agent

nf = (2.5-2)*2 = +1

  1. We multiply the formulae of the oxidizing and reducing agents by suitable integers so as to equalize the total increase or decrease in oxidation number as calculated in steps 3.

Cr2O72- + 2 S2O32- ⟶ 2 Cr3+ + S4O62-

  1. Balance all atoms other than O and H atoms.

Cross multiply the oxidizing and reducing agent with n-factor (nf).

Cr2O72- + 6 S2O32- ⟶ 2 Cr3+ + 3 S4O62-

6.For the ionic reaction in acidic medium case , balance the O-atom by adding H2O in O-atom deficient side.

Cr2O72- + 6 S2O32- ⟶ 2 Cr3+ + 3 S4O62- + 7 H2O

  1. For the ionic reaction in acidic medium case, balance the H-atom by adding H+ ions to side deficient in H-atoms .

Cr2O72- + 6 S2O32- + 14 H+ ⟶ 2 Cr3+ + 3 S4O62- + 7H2O

This is the overall balanced equation by oxidation number method in acidic medium.

 

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