Class 11 Chemistry Chapter 7: Redox Reactions worksheet with answer key

Answer: (c) ZnO → Zn, on heating zinc oxide with carbon

Removal of oxygen from ZnO is reduction. Adding chlorine to FeCl₂, removing hydrogen from ethanol and adding oxygen to copper are all oxidations in the classical sense.

Answer: (c) NH₄Cl(s) → NH₃(g) + HCl(g)

In NH₄Cl → NH₃ + HCl, N stays −3, H +1 and Cl −1, and no element is formed. Ag₂O gives Ag (+1 → 0) and O₂ (−2 → 0); NaNO₃ gives N +5 → +3; ammonium dichromate gives N −3 → 0 and Cr +6 → +3.

Answer: (b) V₂(V)O₅

Each V is +5 (2x − 10 = 0), and Stock notation puts that number as a Roman numeral after the metal's symbol: V₂(V)O₅. X would be the total for both atoms; II is the subscript, not an oxidation number.

Answer: (d) MgH₂

Hydrogen is −1 only when bonded to a metal in a binary compound, as in the hydride MgH₂. In H₂S, NH₄⁺ and HF it is bonded to non-metals and is +1.

Answer: (a) accepts electrons and is itself reduced

An oxidising agent is an acceptor of electrons; gain of electrons is reduction, so the oxidant is itself reduced. A donor of electrons is a reducing agent.

MnO₄⁻ is intensely coloured, so it is its own indicator: the first lasting pink tinge after the last reductant is used up marks the end point. Dichromate shows no such sharp colour change, so diphenylamine is added; Cr₂O₇²⁻ oxidises it just after the equivalence point to an intense blue.

• permanganate is a self indicator (first lasting pink) (1)
• dichromate needs diphenylamine, which turns intense blue (1)

(i) HOCl: (+1) + x + (−2) = 0, so Cl = +1. (ii) Cl₂O₇: 2x + 7(−2) = 0, so Cl = +7. Chlorine is −1 only as a halide ion; combined with oxygen, which is more electronegative, it takes positive oxidation numbers, as in its oxoacids and oxides.

• Cl = +1 in HOCl and +7 in Cl₂O₇, with the sums shown (1)
• positive because chlorine is bonded to the more electronegative oxygen (1)

N goes from 0 in N₂ to −3 in NH₃ (reduced); H goes from 0 in H₂ to +1 in NH₃ (oxidised). NH₃ is covalent and the electrons are only shifted towards nitrogen, but for book-keeping the bonding electrons are taken as wholly transferred to the more electronegative atom, N. Since oxidation numbers change, the reaction is redox.

• N 0 → −3, reduced; H 0 → +1, oxidised (1)
• electron shift treated as complete transfer to the more electronegative N, for book-keeping (1)

(i) Right: one compound breaks into elements, Ag +1 → 0 and O −2 → 0, so it is a redox decomposition. (ii) Wrong: Al 0 → +3 and H +1 → 0; aluminium displaces hydrogen, a non-metal, so it is a hydrogen displacement, not a metal displacement. (iii) Wrong: Cl 0 → −1 and Br −1 → 0 are two different elements. Disproportionation needs one element in one state to be both oxidised and reduced; this is a halogen displacement, chlorine displacing bromide.

• (i) right, with Ag +1 → 0 and O −2 → 0 (1)
• (ii) wrong: hydrogen (non-metal) displacement, Al 0 → +3, H +1 → 0 (1)
• (iii) wrong: two different elements change, so halogen displacement, not disproportionation (1)

Oxidation half: Cr(OH)₃ + 5OH⁻ → CrO₄²⁻ + 4H₂O + 3e⁻ (Cr +3 → +6). Reduction half: IO₃⁻ + 3H₂O + 6e⁻ → I⁻ + 6OH⁻ (I +5 → −1). Multiply the oxidation half by 2 and add: 2Cr(OH)₃ + 10OH⁻ + IO₃⁻ + 3H₂O → 2CrO₄²⁻ + 8H₂O + I⁻ + 6OH⁻. Cancel: 2Cr(OH)₃(s) + IO₃⁻(aq) + 4OH⁻(aq) → 2CrO₄²⁻(aq) + I⁻(aq) + 5H₂O(l). Check: O 13 = 13, H 10 = 10, charge −5 = −5.

• oxidation half balanced in base with 3e⁻ (1)
• reduction half balanced in base with 6e⁻ (1)
• halves added in 2 : 1, cancelled to the final equation (1)

Al₂S₃ is ionic, (Al³⁺)₂(S²⁻)₃. Oxidation half: 2Al → 2Al³⁺ + 6e⁻. Reduction half: 3S + 6e⁻ → 3S²⁻. Their sum gives the overall reaction. Aluminium donates electrons, so it is the reducing agent; sulphur accepts them, so it is the oxidising agent.

• oxidation half 2Al → 2Al³⁺ + 6e⁻ (1)
• reduction half 3S + 6e⁻ → 3S²⁻ (1)
• reductant Al, oxidant S (1)