Question

Calculate the oxidation number of chlorine in the following redox reaction and determine how many moles of water are involved in the balanced equation for this reaction in basic conditions.:

ClO₂^1- (aq) + N₂H₄ (g) ↔ NO (g) + Cl₂ (g)

Answer 1

ClO21- (aq) + N2H4 (g) ↔ NO (g) + Cl2 (g)

For chlorine...

initially :

ClO2(-) that is, the overall molecule is -1 charged

Oxygen will typically have a -2 charge

so

Cl + 2(-2) = -1

Cl = -1+4 = +3

Cl is +3 charged

in the products, Cl is Cl2 present, meaning it has 0 in oxidation number

therefore, +3 --> 0

meaning 2Cl gains + 2(3e-) to form Cl2(g)

oxidation number is +3, then 0 in product

Q2.

moles of water involved in the balanced  reaction in basic conditions

split in half cells:

ClO2- = Cl2

N2H4 = NO2

1, balance atoms other than O,H

2ClO2- = Cl2

N2H4 = 2NO2

2, balance O, adding H2O

2ClO2- = Cl2 + 4H2O

4H2O + N2H4 = 2NO2

3, balance H adding H+

8H+ + 2ClO2- = Cl2 + 4H2O

4H2O + N2H4 = 2NO2 + 8H+

4, balance charges

6e- + 8H+ + 2ClO2- = Cl2 + 4H2O

4H2O + N2H4 = 2NO2 + 8H+ + 8e-

5, balance electrons

24e- + 32H+ + 8ClO2- = 4Cl2 + 16H2O

12H2O + 3N2H4 = 6NO2 + 24H+ + 24e-

6, add all

24e- + 32H+ + 8ClO2- + 12H2O + 3N2H4  = 4Cl2 + 16H2O + 6NO2 + 24H+ + 24e-

7, cancel common terms

8H+ + 8ClO2- + 3N2H4  = 4Cl2 + 4H2O + 6NO2

8., add OH- for basic conditions

8OH+ + 8H+ + 8ClO2- + 3N2H4  = 4Cl2 + 4H2O + 6NO2 + 8OH-

9, add all H+ and OH-

8H2O + 8ClO2- + 3N2H4  = 4Cl2 + 4H2O + 6NO2 + 8OH-

10, cancel common terms

4H2O + 8ClO2- + 3N2H4  = 4Cl2 + 6NO2 + 8OH-

water moles = 4