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## 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.

## 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) + 4$OH^{-}$(aq) reactants: -1 + 3(-1) = -4       products: 4(-1) = -4 (e) Balance charge reactants: 0 H       products: 4 H To balance, add 2$H_{2}O$ on the left. $NO_{3}^{-}(aq)$ + $3e^{-}$ + 2$H_{2}O$ → NO(g) + 4$OH^{-}(aq)$ reactants: 4 H       products: 4 H (f) Balance H reactants: 3 + 2 = 5       products: 4 + 1 = 5 $NO_{3}^{-}(aq)$ + $3e^{-}$  → NO(g) + 4$OH^{-}$(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. $Cl_{2}(g)$ → 2$ClO_{3} ^{-}(aq)$ (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 $Cl_{2}(g)$ → 2$ClO_{3} ^{-}(aq)$ + 10$e^{-}$ (d) Add electrons reactants: 0       products: 2(–1) + 10(–1) = –12 acidic medium: add $H^{+}$. To balance, add 12 $H^{+}$ on the right. $Cl_{2}(g)$ → 2$ClO_{3} ^{-}(aq)$ + 10$e^{-}$ + 12$H^{+}$(aq) reactants: 0 products: 2(-1) + 10(-1) + 12(+1) = 0 (e) Balance charge reactants: 0 H    products: 12 H To balance, add 6$H_{2}O$ on the left. $Cl_{2}(g)$ + 6$H_{2}O$ → 2$ClO_{3} ^{-}(aq)$ + 10$e^{-}$ + 12$H^{+}$(aq) reactants: 12 H       products: 12 H (f) Balance H reactants: 6       products: 2(3) = 6 $Cl_{2}(g)$ + 6$H_{2}O$ → 2$ClO_{3} ^{-}(aq)$ + 10$e^{-}$ + 12$H^{+}$(aq) (g) Check O: The half-equation is balanced: