Subscribe $4.99/month

Un-lock Verified Step-by-Step Experts Answers.

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

Our Website is free to use.
To help us grow, you can support our team with a Small Tip.

All the data tables that you may search for.

Need Help? We got you covered.

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Products

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Need Help? We got you covered.

Chapter 4

Q. 4.8

Balance the following half-equations:

(1)NO_{3}^{-}(aq) → NO(g) (basic solution)

(2)Cl_{2}(g)ClO_{3} ^{-}(aq) (acidic solution)

STRATEGY

Follow the steps outlined above in the order given.

Step-by-Step

Verified Solution

N: +5 → +2; O: -2 → -2; N is reduced. (1) (a) Oxidation numbers
1 N on each side; no adjustment is required. (b) Atom balance
N: 5(1) → N: 2(1) (c) “total oxidation” number
The oxidation number for N goes from +5 to +2. It is reduced by 3.

Add 3 electrons to the reactant side.

NO_{3}^{-}(aq) + 3e^{-}  → NO(g)

(d) Add electrons
reactants: -1 + 3(-1) = -4       products: 0

basic medium: add OH^{-}. To balance, add 4 OH^{-} on the right.

NO_{3}^{-}(aq) + 3e^{-}  → NO(g) + 4OH^{-}(aq)

reactants: -1 + 3(-1) = -4       products: 4(-1) = -4

(e) Balance charge
reactants: 0 H       products: 4 H
To balance, add 2H_{2}O on the left.

NO_{3}^{-}(aq) + 3e^{-} + 2H_{2}O → NO(g) + 4OH^{-}(aq)

reactants: 4 H       products: 4 H

(f) Balance H
reactants: 3 + 2 = 5       products: 4 + 1 = 5

NO_{3}^{-}(aq) + 3e^{-}  → NO(g) + 4OH^{-}(aq)

(g) Check O:

The half-equation is balanced:

Cl: 0 → +5; O: -2 → -2; Cl is oxidized. (2) (a) Oxidation numbers
reactant: 2 Cl       product: 1 Cl
Multiply ClO_{3}^{-} by 2.
Cl_{2}(g) → 2ClO_{3} ^{-}(aq)
(b) Atom balance
Cl: 0(2) = 0 → Cl: 5(2) = 10 (c) “total oxidation” number
The oxidation number for Cl goes from 0 to 10. The oxidation number increases
by 10. Add 10 electrons to the product side

Cl_{2}(g) → 2ClO_{3} ^{-}(aq) + 10e^{-}

(d) Add electrons
reactants: 0       products: 2(–1) + 10(–1) = –12

acidic medium: add H^{+}. To balance, add 12 H^{+} on the right.

Cl_{2}(g) → 2ClO_{3} ^{-}(aq) + 10e^{-} + 12H^{+}(aq)

reactants: 0 products: 2(-1) + 10(-1) + 12(+1) = 0

(e) Balance charge
reactants: 0 H    products: 12 H

To balance, add 6H_{2}O on the left.

Cl_{2}(g) + 6H_{2}O → 2ClO_{3} ^{-}(aq) + 10e^{-} + 12H^{+}(aq)

reactants: 12 H       products: 12 H

(f) Balance H
reactants: 6       products: 2(3) = 6

Cl_{2}(g) + 6H_{2}O → 2ClO_{3} ^{-}(aq) + 10e^{-} + 12H^{+}(aq)

(g) Check O:

The half-equation is balanced: