Given the partial equation ? mno4−+ ? so32− → ? mn2++ ? so42− what must the coefficients be so that the electrons are balanced? enter the equation coefficients in order separated by commas (e. g., 2,2,1,4, where 1 indicates the absence of a coefficient).

The coefficients should be: 2, 5, 2, 5

Explanation:

Given redox reaction: MnO₄⁻ + SO₃²⁻ → Mn²⁺+ SO₄²⁻

To balance the given redox reaction in acidic medium, the oxidation and the reduction half-reactions should be balanced first.

Reduction half-reaction: MnO₄⁻ → Mn²⁺

Oxidation state of Mn in MnO₄⁻ is +7 and the oxidation state of Mn in Mn²⁺ is +2. Therefore, Mn accepts 5e⁻ to get reduced from +7 to +2 oxidation state.

⇒ MnO₄⁻ + 5e⁻ → Mn²⁺

Now the total charge on reactant side is (-6) and the total charge on product side is +2. Therefore, to balance the total charge, 8H⁺ must be added to the reactant side.

⇒ MnO₄⁻ + 5e⁻ + 8H⁺ → Mn²⁺

To balance the number of hydrogen and oxygen atoms, 4H₂O must be added to the product side.

⇒ MnO₄⁻ + 5e⁻ + 8H⁺ → Mn²⁺ + 4H₂O               equation 1

Oxidation half-reaction: SO₃²⁻ → SO₄²⁻

Oxidation state of S in SO₃²⁻ is +4 and the oxidation state of S in SO₄²⁻ is +6. Therefore, S loses 2e⁻ to get oxidized from +4 to +6 oxidation state.

⇒ SO₃²⁻ → SO₄²⁻ + 2e⁻

Now the total charge on reactant side is (-2) and the total charge on product side is (-4). Therefore, to balance the total charge, 2H⁺ must be added to the product side.

⇒ SO₃²⁻ → SO₄²⁻ + 2e⁻ + 2H⁺

To balance the number of hydrogen and oxygen atoms, 1 H₂O must be added to the reactant side.

⇒ SO₃²⁻ + H₂O → SO₄²⁻ + 2e⁻ + 2H⁺                equation 2

Now, to cancel the electrons transferred, equation (1) is multiplied by 2 and equation (2) is multiplied by 5.

Balanced Reduction half-reaction:

MnO₄⁻ + 5e⁻ + 8H⁺ → Mn²⁺ + 4H₂O ] × 2

⇒ 2MnO₄⁻ + 10e⁻ + 16H⁺ → 2Mn²⁺ + 8H₂O                equation 3

Balanced Oxidation half-reaction:

SO₃²⁻ + H₂O → SO₄²⁻ + 2e⁻ + 2H⁺ ] × 5

⇒ 5SO₃²⁻ + 5H₂O → 5SO₄²⁻ + 10e⁻ + 10H⁺                  equation 4  

Now adding equation 3 and 4, to obtain the overall balanced redox reaction:

2MnO₄⁻ + 5SO₃²⁻ + 6H⁺ → 2Mn²⁺ + 5SO₄²⁻ + 3H₂O

Therefore, the coefficients should be: 2, 5, 2, 5

false

because it does not have the potential to be entirely used up

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