Balance each of the following redox reactions occurring in basic solution.

1) h2o2(aq)+clo2(aq)→clo−2(aq)+o2(g)
2) al(s)+mno−4(aq)→mno2(s)+al(oh)−4(aq )

3) cl2(g)→cl−(aq)+clo−(aq)

1) H₂O₂(aq) + 2OH⁻(aq) + 2ClO₂(aq) → 2ClO₂⁻(aq) + 2H₂O(l) + O₂(g)

2) Al(s) + MnO₄⁻(aq) + 2H₂O(l)→ Al(OH)₄⁻(aq) + MnO₂(s)

3) 2Cl₂(g) + 2OH⁻(aq) → 2Cl⁻(aq) + → 2ClO⁻(aq) + 2H⁺(aq)

Explanation:

To balance a redox reaction we need to find out what substance is being reduced and what substance is being oxidized. The reduction occurs when the oxidation number (nox) decreases, it happens when the atom gains electrons.

On the other hand, the oxidation happens when the oxidation number increases, it means that the atom is losing electrons. Thus, let's calculate the nox of the atoms in each substance and proceed with the balancement.

1) H₂O₂(aq) + ClO₂(aq) → ClO₂⁻(aq) + O₂(g)

H₂O₂: H: fixed nox + 1, O: 2x + 2*1 = 0 -> x = -1

ClO₂: O: fixed nox -1 (peroxide), Cl: x +2*(-1) = 0 -> x = +2

ClO₂⁻: O: fixed nox -2, Cl: x + (-2) = -1 -> x = +1

O₂: O: simple substane, nox = 0.

Thus, H₂O₂ oxides and ClO₂ reduces.

Because it is at a basic solution, there will be a OH⁻ being reduced too:

H₂O₂(aq) + OH⁻(aq) → H₂O(l) + O₂(g) + 2e⁻

ClO₂ (aq) + e⁻ → ClO₂⁻(aq)

The equations must have the same number of each element on both sides, and the same number of electrons. So, in the first, we multiply OH⁻ and H₂O by 2, and the second equation by 2:

H₂O₂(aq) + 2OH⁻(aq) → 2H₂O(l) + O₂(g) + 2e⁻

2ClO₂ (aq) + 2e⁻ → 2ClO₂⁻(aq)

Summing the equations:

H₂O₂(aq) + 2OH⁻(aq) + 2ClO₂(aq) → 2ClO₂⁻(aq) + 2H₂O(l) + O₂(g)

2) Al(s) + MnO₄⁻(aq) → MnO₂(s) + Al(OH)₄⁻(aq)

Al: simple element, nox = 0

MnO₄⁻: O: nox fix = -2, Mn: x + 4*(-2) = -1 -> x = +7

MnO₂: O: nox fix = -2, Mn: x + 2*(-2) = 0 -> x = +4

Al(OH)₄⁻: OH: nox fix = -1, Al: x + 4*(-1) = -1 -> x = +3

Al oxides in the presence of the OH⁻ ions, and MnO₄⁻ reduces in the presence of water as the half reactions:

Al(s) + OH⁻(aq) → Al(OH)₄⁻(aq) + 3e⁻

MnO₄⁻(aq) + H₂O(l) + 3e⁻ → MnO₂(s) + OH⁻(aq)

The number of electrons are already balanced, but at the second equation the number of oxygens and hydrogens are not. So, we multiply OH⁻ by 4 and H₂O by 2. And the first equation we multiply OH⁻ by 4:

Al(s) + 4OH⁻(aq) → Al(OH)₄⁻(aq) + 3e⁻

MnO₄⁻(aq) + 2H₂O(l) + 3e⁻ → MnO₂(s) + 4OH⁻(aq)

Summing the equations:

Al(s) + OH⁻(aq) + MnO₄⁻(aq) + 2H₂O(l)→ Al(OH)₄⁻(aq) + MnO₂(s) + 4OH⁻(aq)

Simplifying OH⁻ :

Al(s) + MnO₄⁻(aq) + 2H₂O(l)→ Al(OH)₄⁻(aq) + MnO₂(s)

3) Cl₂(g) → Cl⁻(aq) + ClO⁻(aq)

Cl₂: Cl: simple compound, nox 0

Cl⁻: Cl: ion, nox: -1

ClO⁻: O: fix nox -2, Cl: x -2 = -1 -> x = +1

Thus, Cl₂ oxides and reduces at same time. The half reactions are:

Cl₂(g) + 2e⁻→ Cl⁻(aq)

Cl₂(g) + OH⁻(aq) → ClO⁻(aq) + H⁺(aq) + 2e⁻

In the first equation we multiply Cl⁻ by 2, and in the second, ClO⁻, OH⁻, and H⁺ by 2:

Cl₂(g) + 2e⁻→ 2Cl⁻(aq)

Cl₂(g) + 2OH⁻(aq) → 2ClO⁻(aq) + 2H⁺(aq) + 2e⁻

Summing the equations:

2Cl₂(g) + 2OH⁻(aq) → 2Cl⁻(aq) + → 2ClO⁻(aq) + 2H⁺(aq)

the answers would be

the first one and the second one

a electronegativity decreases beacuse the size of nucleus is bigger.