Question

9. What are the coefficients in front of H* and H20 after balancing of the redox equation in acidic conditions? BizO3(s) + OCH(aq) → CH(aq) + Bios"(aq) A. H = 0; H20-1 B. H = 1; H20-2 C. H = 2; H20-1 +6 Hr6e- D. H = 2; H20 - 2 E. H* = 6; H20 - 3 Big Og 3460 > 2B, 0 Buracioci co- +1400)

Answer 1

ANSWER : C) H ⁺ = 2 &  H₂O = 1

Consider given reaction, Bi₂O₃ (s) + OCl ^- (aq) $\rightarrow$ Cl ^- (aq) + BiO₃^- (aq)

Oxidation no. of Bi changes from +3 to +5 , hence it is oxidized. We can write oxidation half equation as

Bi₂O₃ (s) $\rightarrow$ BiO₃^- (aq)

Oxidation no. of Cl changes from +1 to -1 , hence Cl is reduced. We can write reduction half equation as

OCl ^- (aq) $\rightarrow$ Cl ^- (aq)

Now, balance the half equations for atoms other than O and H.

Oxidation : Bi₂O₃ (s) $\rightarrow$ 2 BiO₃^- (aq)

Reduction : OCl ^- (aq) $\rightarrow$ Cl ^- (aq)

Balance half equations for O atoms by adding H₂O to the side of half equations with less number of O atoms.

Oxidation : Bi₂O₃ (s) + 3 H₂O (l)  $\rightarrow$2 BiO₃^- (aq)

Reduction : OCl ^- (aq) $\rightarrow$ Cl ^- (aq) + H₂O (l)

Balance half equations for H atoms by adding H ⁺ to the side of half equations with less number of H atoms.

Oxidation : Bi₂O₃ (s) + 3 H₂O (l)  $\rightarrow$2 BiO₃^- (aq) + 6 H ⁺ (aq)

Reduction : OCl ^- (aq) + 2 H ⁺ (aq)  $\rightarrow$ Cl ^- (aq) + H₂O (l)

Balance half equations for charge by adding electrons. Charge on L.H.S must be equal to R.H.S of half equation.

Oxidation : Bi₂O₃ (s) + 3 H₂O (l)  $\rightarrow$2 BiO₃^- (aq) + 6 H ⁺ (aq) + 4 e ^-

Reduction : OCl ^- (aq) + 2 H ⁺ (aq) + 2 e ^-$\rightarrow$ Cl ^- (aq) + H₂O (l)

To make electrons equal in both equation, multiply reduction half equation by 2.

Oxidation : Bi₂O₃ (s) + 3 H₂O (l)  $\rightarrow$2 BiO₃^- (aq) + 6 H ⁺ (aq) + 4 e ^-

Reduction : 2 OCl ^- (aq) + 4 H ⁺ (aq) + 4 e ^-$\rightarrow$ 2 Cl ^- (aq) + 2 H₂O (l)

Add two equations to get overall equation.

Bi₂O₃ (s) + 3 H₂O (l)  $\rightarrow$2 BiO₃^- (aq) + 6 H ⁺ (aq) + 4 e ^-

2 OCl ^- (aq) + 4 H ⁺ (aq) + 4 e ^-$\rightarrow$ 2 Cl ^- (aq) + 2 H₂O (l)

Overall : Bi₂O₃ (s) +  2 OCl ^- (aq) + H₂O (l)  $\rightarrow$2 BiO₃^- (aq) + 2 Cl ^- (aq) + 2 H ⁺ (aq)

ANSWER : H ⁺ = 2 &  H₂O = 1