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Travka [436]
3 years ago
6

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:

<em>5 H₂O</em> + As₂O₃(s) + 4 e⁻ +<u> 6H⁺</u> + 2 NO₃⁻ (aq) → 2 H₃AsO₄(aq)  +<u> 4 H⁺</u> + 4 e⁻ + N₂O₃(aq)  + <em>3 H₂O</em>  

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

I hope it helps!

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