Redox Chemistry

Drow D[1]

FeTiO₃: +2 + x − 6 = 0, so x = +4.

Fe₂(TiO₃)₃: 2(+3) + 3x − 18 = 0, so x = +4.

Acarbon dioxide and calcium carbonate[1]

Carbon is +4 in CO₂ and in CaCO₃.

In KCN (K +1, N −3) carbon is +2, and in Fe₂C (Fe +2) it is −4.

C+1, +2, +3[1]

Bi, Sr, Ca and O give 6 + 4 + 4 − 20 = −6, so the three copper atoms total +6.

Only +1, +2, +3 adds up to +6.

Dtwo oxidations, highest +6[1]

Sulfur goes 0, +4, +6, +6, +6: only the first two stages raise it.

In H₂S₂O₇, 2 + 2x − 14 = 0 gives +6 for each sulfur, not +12.

Crow C[1]

Sulfur starts at +6. In reaction 1 it stays +6 (0). In reaction 2 it falls to +4 in SO₂ (2).

In reaction 3 it falls to −2 in H₂S (8).

Bnitrogen from +4 to 0[1]

In NO nitrogen is +2, not +4, so it changes from +2 to 0.

Carbon goes from +2 to +4 and oxygen stays −2, so the reaction is redox.

COxygen is reduced in stage 1.[1]

In stage 1 oxygen falls from 0 to −2, so it is reduced, while nitrogen rises from +2 to +4.

In stage 2 NO₂ (N +4 to +2) is the oxidising agent and SO₂ (S +4 to +6) the reducing agent.

C1, 2 and 4[1]

Sulfur dioxide is an acidic oxide: it does not react with an acid but forms a salt with an alkali. It is a reducing agent, so it decolourises acidified manganate(VII).

Its solution is acidic, so Universal Indicator turns red or orange, not violet.

B1 and 3 only[1]

2FeCl₃ + 2KI → 2FeCl₂ + 2KCl + I₂, so v = w = x = y = 2, and KI is the test for an oxidising agent.

The iodine stays dissolved, giving a yellow-brown solution, not a vapour.

B2 and 4[1]

Nitrogen is +3 in NF₃, and each F₂ goes from 0 to two F⁻, gaining 2 electrons.

Nitrogen is only oxidised (it stays −3 in NH₄F), so there is no disproportionation, and F₂ is the oxidising agent.

C+3[1]

n_ClO₄⁻ = 5.00 × 10⁻³ mol, n_Ti³⁺ = 0.0200 mol, each losing 1 electron.

Each Cl gains 0.0200 ÷ 5.00 × 10⁻³ = 4 electrons: +7 − 4 = +3.

CAmO₂²⁺[1]

n_Am³⁺ = 2.40 × 10⁻⁴ mol and n_Cu³⁺ = 7.20 × 10⁻⁴ mol, each gaining 1 electron, so each Am loses 3: +3 to +6.

In AmO₂²⁺, x − 4 = +2 gives +6. The others give +3, +4 and +3.

C+4[1]

M starts at +5 in M₂O₅, and n_M = 2 × 2.50 × 10⁻³ = 5.00 × 10⁻³ mol.

n_MnO₄⁻ = 1.00 × 10⁻³ mol takes 5.00 × 10⁻³ mol of electrons: 1 per M, so M was at +4.

B40.0[1]

n_MnO₄⁻ = 3.00 × 10⁻³ mol takes 0.0150 mol of electrons, so n_M²⁺ = 0.0150 mol = n_MO.

M_r(MO) = 0.84 ÷ 0.0150 = 56.0, so A_r(M) = 56.0 − 16.0 = 40.0. Option D forgets the oxygen.

C+5[1]

0.2 mol of I⁻ gives 0.10 mol of I₂ and loses 0.2 mol of electrons. The other 0.02 mol of I₂ (0.04 mol of I atoms) comes from the oxide.

0.04 mol of I atoms gain 0.2 mol of electrons: 5 each, so the oxide iodine is +5.

C+3[1]

n_FeC₂O₄ = 4.0 × 10⁻⁴ mol, each giving 1 + 2 = 3 electrons: 1.2 × 10⁻³ mol.

n_MnO₄⁻ = 3.0 × 10⁻⁴ mol, so each Mn gains 4: +7 − 4 = +3.

ASO₂[1]

2 mol of H₂S (S −2) lose 4 mol of electrons in reaching 0.

So 1 mol of the other sulfur must gain 4 electrons: it starts at +4, as in SO₂.

B0.150 mol[1]

Cl +5 to +7 loses 2 electrons. Cl +5 to −1 gains 6. So three chlorate(VII) form for each chloride: 4KClO₃ → 3KClO₄ + KCl.

0.200 × 3/4 = 0.150 mol.

B0.133 mol/dm³[1]

n_{S₂O₃²⁻} = 2.0 × 10⁻³ mol, so n_I₂ = 1.0 × 10⁻³ mol and n_Cu²⁺ = 2.0 × 10⁻³ mol.

c = 2.0 × 10⁻³ ÷ 0.015 = 0.133 mol/dm³.

B6.40 mg/dm³[1]

4.00 × 10⁻³ mol S₂O₃²⁻ needs 2.00 × 10⁻³ mol I₂, from 2.00 × 10⁻³ mol MnO(OH)₂, from 1.00 × 10⁻³ mol O₂.

1.00 × 10⁻³ × 32.0 = 0.0320 g = 32.0 mg in 5.00 dm³, so 6.40 mg/dm³.

B0.050[1]

n_BaSO₄ = 0.583 ÷ 233.4 = 2.50 × 10⁻³ mol in 25.0 cm³, so 0.0250 mol of sulfate in 250 cm³.

Each S₂O₃²⁻ gives 2SO₄²⁻, so n_{S₂O₃²⁻} = 0.0125 mol in 0.250 dm³: 0.050 mol dm⁻³.

A0.0440[1]

n_{S₂O₃²⁻} = 2.20 × 10⁻³ mol, so n_I₂ = 1.10 × 10⁻³ mol. 3I₂ needs 6H⁺, so n_H⁺ = 2.20 × 10⁻³ mol.

H₂SO₄ gives 2H⁺: 1.10 × 10⁻³ mol in 0.0250 dm³ = 0.0440 mol dm⁻³.

C2.80[1]

n_{S₂O₃²⁻} = 4.0 × 10⁻³ mol, n_I₂ = 2.0 × 10⁻³ mol, n_CO = 0.010 mol.

0.010 × 28.0 = 0.280 g in 0.100 dm³ = 2.80 g dm⁻³.

B40 cm³[1]

Each formula unit contains two C₂O₄ groups, so 2.0 × 10⁻³ mol is oxidised.

n_MnO₄⁻ = 2/5 × 2.0 × 10⁻³ = 8.0 × 10⁻⁴ mol, V = 8.0 × 10⁻⁴ ÷ 0.020 = 0.040 dm³.

B11.4 cm³[1]

2MnO₄⁻ + 5H₂C₂O₄: 2.5 × 10⁻⁴ mol of each, so the acid is limiting (it needs only 1.0 × 10⁻⁴ mol of MnO₄⁻).

n_CO₂ = 2 × 2.5 × 10⁻⁴ = 5.0 × 10⁻⁴ mol × 22.7 dm³ = 11.4 cm³.