Respuesta :
Answer:
[tex]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)[/tex]
Explanation:
Oxidation: [tex]As_{2}O_{3}(s)\rightarrow H_{3}AsO_{4}(aq)[/tex]
- Balance As: [tex]As_{2}O_{3}(s)\rightarrow 2H_{3}AsO_{4}(aq)[/tex]
- Balance H and O in acidic medium: [tex]As_{2}O_{3}(s)+5H_{2}O(l)\rightarrow 2H_{3}AsO_{4}(aq)+4H^{+}(aq)[/tex]
- Balnce charge: [tex]As_{2}O_{3}(s)+5H_{2}O(l)-4e^{-}\rightarrow 2H_{3}AsO_{4}(aq)+4H^{+}(aq)[/tex]......(1)
Reduction: [tex]NO_{3}^{-}(aq)\rightarrow N_{2}O_{3}(aq)[/tex]
- Balance N: [tex]2NO_{3}^{-}(aq)\rightarrow N_{2}O_{3}(aq)[/tex]
- Balance H and O in acidic medium: [tex]2NO_{3}^{-}(aq)+6H^{+}(aq)\rightarrow N_{2}O_{3}(aq)+3H_{2}O(l)[/tex]
- Balance charge: [tex]2NO_{3}^{-}(aq)+6H^{+}(aq)+4e^{-}\rightarrow N_{2}O_{3}(aq)+3H_{2}O(l)[/tex]......(2)
[tex]Equation (1)+Equation (2)[/tex] gives-
[tex]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)[/tex]
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!
