Balance the following redox equation using the smallest integers possible and select the correct coefficient for the hydrogen sulfite ion. MnO41-(aq) + HSO31-( aq) + H+( aq) → Mn2 ( aq) + SO42–( aq) + H2O( l)

Respuesta :

Answer:

Coefficient of [tex]HSO_{3}^{-}[/tex] is 5

Explanation:

Oxidation: [tex]MnO_{4}^{-}(aq)\rightarrow Mn^{2+}(aq)[/tex]

  • Balance O and H in acidic medium: [tex]MnO_{4}^{-}(aq)+8H^{+}(aq)\rightarrow Mn^{2+}(aq)+4H_{2}O(l)[/tex]
  • Balance charge : [tex]MnO_{4}^{-}(aq)+8H^{+}(aq)+5e^{-}\rightarrow Mn^{2+}(aq)+4H_{2}O(l)[/tex] .............(1)

Reduction: [tex]HSO_{3}^{-}\rightarrow SO_{4}^{2-}[/tex]

  • Balance O and H in acidic medium : [tex]HSO_{3}^{-}(aq)+H_{2}O(l)\rightarrow SO_{4}^{2-}(aq)+3H^{+}(aq)[/tex]
  • Balance charge : [tex]HSO_{3}^{-}(aq)+H_{2}O(l)-2e^{-}\rightarrow SO_{4}^{2-}(aq)+3H^{+}(aq)[/tex] ..............(2)

[tex][2\times eq(1)]+[5\times eq(2)][/tex] -

Balanced equation: [tex]2MnO_{4}^{-}(aq)+5HSO_{3}^{-}(aq)+H^{+}(aq)\rightarrow 2Mn^{2+}(aq)+5SO_{4}^{2-}(aq)+3H_{2}O(l)[/tex]

Coefficient of [tex]HSO_{3}^{-}[/tex] is 5

The balanced redox equation using the smallest integers possible is:

2 MnO₄⁻ + H⁺ + 5 HSO₃⁻ → 2 Mn²⁺ + 3 H₂O + 5 SO₄²⁻

What is a redox reaction?

Redox is a type of chemical reaction in which the oxidation states of atoms are changed.

  • Step 1. Write the unbalanced equation.

MnO₄⁻(aq) + HSO₃⁻(aq) + H⁺(aq) → Mn²⁺(aq) + SO₄²⁻(aq) + H₂O(l)

  • Step 2. Write both half-reactions.

Reduction: MnO₄⁻ → Mn²⁺

Oxidation: HSO₃⁻ → SO₄²⁻

  • Step 3. Perform the mass balance by adding H⁺ and H₂O to the deficient side.

MnO₄⁻ + 8 H⁺ → Mn²⁺ + 4 H₂O

H₂O + HSO₃⁻ → SO₄²⁻ + 3 H⁺

  • Step 4. Perform the charge balance by electrons to the deficient side.

MnO₄⁻ + 8 H⁺ + 5 e⁻ → Mn²⁺ + 4 H₂O

H₂O + HSO₃⁻ → SO₄²⁻ + 3 H⁺ + 2 e⁻

  • Step 5. Make the number of electrons gained and lost equal. Sum both half-reactions and cancel what is repeated in both sides.

2 × (MnO₄⁻ + 8 H⁺ + 5 e⁻ → Mn²⁺ + 4 H₂O)

5 × (H₂O + HSO₃⁻ → SO₄²⁻ + 3 H⁺ + 2 e⁻)

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2 MnO₄⁻ + 16 H⁺ + 10 e⁻ + 5 H₂O + 5 HSO₃⁻ → 2 Mn²⁺ + 8 H₂O + 5 SO₄²⁻ + 15 H⁺ + 10 e⁻

2 MnO₄⁻ + H⁺ + 5 HSO₃⁻ → 2 Mn²⁺ + 3 H₂O + 5 SO₄²⁻

The balanced redox equation using the smallest integers possible is:

2 MnO₄⁻ + H⁺ + 5 HSO₃⁻ → 2 Mn²⁺ + 3 H₂O + 5 SO₄²⁻

Learn more about redox equations here: https://brainly.com/question/21851295

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