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3 years ago

Complete and balance the following redox reaction in acidic solution As2O3(s) + NO3- (aq) → H3AsO4(aq) + N2O3(aq)

Chemistry

2 answers:

scoray [572]3 years ago

3 0

Answer:

$As_{2}O_{3}(s)+2NO_{3}^{-}(aq)+2H_{2}O(l)+2H^{+}(aq)\rightarrow 2H_{3}AsO_{4}(aq)+N_{2}O_{3}(aq)$

Explanation:

Oxidation: $As_{2}O_{3}(s)\rightarrow H_{3}AsO_{4}(aq)$

Balance As: $As_{2}O_{3}(s)\rightarrow 2H_{3}AsO_{4}(aq)$
Balance H and O in acidic medium: $As_{2}O_{3}(s)+5H_{2}O(l)\rightarrow 2H_{3}AsO_{4}(aq)+4H^{+}(aq)$
Balnce charge: $As_{2}O_{3}(s)+5H_{2}O(l)-4e^{-}\rightarrow 2H_{3}AsO_{4}(aq)+4H^{+}(aq)$ ......(1)

Reduction: $NO_{3}^{-}(aq)\rightarrow N_{2}O_{3}(aq)$

Balance N: $2NO_{3}^{-}(aq)\rightarrow N_{2}O_{3}(aq)$
Balance H and O in acidic medium: $2NO_{3}^{-}(aq)+6H^{+}(aq)\rightarrow N_{2}O_{3}(aq)+3H_{2}O(l)$
Balance charge: $2NO_{3}^{-}(aq)+6H^{+}(aq)+4e^{-}\rightarrow N_{2}O_{3}(aq)+3H_{2}O(l)$ ......(2)

$Equation (1)+Equation (2)$ gives-

$As_{2}O_{3}(s)+2NO_{3}^{-}(aq)+2H_{2}O(l)+2H^{+}(aq)\rightarrow 2H_{3}AsO_{4}(aq)+N_{2}O_{3}(aq)$

Maurinko [17]3 years ago

3 0

Answer:

2 H₂O + As₂O₃(s) + 2 H⁺ + 2 NO₃⁻ (aq) → 2 H₃AsO₄(aq) + N₂O₃(aq)

Explanation:

As₂O₃(s) + NO₃⁻ (aq) → H₃AsO₄(aq) + N₂O₃(aq)

To balance a redox reaction the first step is to know which atom is oxidated and which reduced. To know it. We need to obtain the oxidation number for all atoms.

There is a rule. All hydrogens have oxidation number +1 and all oxygens -2. I will show how to calculate oxidation numbers of Nitrogen and Arsenic.

For As₂O₃: 3 oxygens are -6 and total oxidation number of this compound is 0. So, to balances charges two arsenic must have +3 of oxidation number:

0 = -2 × 3 + +3 × 2

Oxidation number of As₂O₃ = oxygens + Arsenics

For NO₃⁻: Three oxygens are -6 and total oxidation is -1. So, Nitrogen is +5:

-1 = -2×3 + +5 × 1

Oxidation number of NO₃⁻ = oxygens + Nitrogen

Thus, oxidation numbers are:

As₂O₃(s) + NO₃⁻ (aq) → H₃AsO₄(aq) + N₂O₃(aq)

+3 +5 +5 +3

The atoms which increase oxidation number is oxidated and the atoms which decrease oxidation number is reduced. Thus, As is oxidated and N is reduced.

The next step is separate half-reactions, thus:

As₂O₃(s) → H₃AsO₄(aq) Oxidation

NO₃⁻ (aq) → N₂O₃(aq) Reduction

Then, we should balance, first, elements differents of oxygen and hydrogen:

As₂O₃(s) → 2 H₃AsO₄(aq) Oxidation

2 NO₃⁻ (aq) → N₂O₃(aq) Reduction

Then, balance oxygens with H₂O and hydrogens with H⁺ (Because is acidic solution):

5 H₂O + As₂O₃(s) → 2 H₃AsO₄(aq) + 4 H⁺ Oxidation

6H⁺ + 2 NO₃⁻ (aq) → N₂O₃(aq) + 3 H₂O Reduction

After this, we must balance charges with electrons:

5 H₂O + As₂O₃(s) → 2 H₃AsO₄(aq) + 4 H⁺ + 4 e⁻ Oxidation

4 e⁻ + 6H⁺ + 2 NO₃⁻ (aq) → N₂O₃(aq) + 3 H₂O Reduction

Electron number must be the same for oxidation and reduction
Last, we should sum half-reactions and cancel out common compounds:

5 H₂O + As₂O₃(s) + 4 e⁻ + 6H⁺ + 2 NO₃⁻ (aq) → 2 H₃AsO₄(aq) + 4 H⁺ + 4 e⁻ + N₂O₃(aq) + 3 H₂O

2 H₂O + As₂O₃(s) + 2 H⁺ + 2 NO₃⁻ (aq) → 2 H₃AsO₄(aq) + N₂O₃(aq)

I hope it helps!

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