CBSE Class 11 Chemistry Redox Reactions Worksheet Set 04

Redox Reactions Worksheet

Question 1. Name the following compounds using Stock notation: Cu₂O, FeO, SnCl₄, Mn₂O₃, Hg₂Cl₂.
Answer:
Cu₂O: 2x − 2 = 0, x = +1, copper(I) oxide.
FeO: iron(II) oxide.
SnCl₄: Sn = +4, tin(IV) chloride.
Mn₂O₃: 2x = +6, manganese(III) oxide.
Hg₂Cl₂: each Hg = +1, mercury(I) chloride.
In simple words: Find the metal's oxidation number from the anion and write it in Roman numerals in brackets.

Teacher's Note:
1. Do not put a space before the bracket in the formula form, e.g. Sn(IV)Cl₄.
2. Hg₂²⁺ is one ion containing two Hg atoms, each +1.

Question 2. Classify each as combination, decomposition or displacement and say whether it is redox: (a) C + O₂ → CO₂ (b) 2HgO → 2Hg + O₂ (c) Zn + CuSO₄ → ZnSO₄ + Cu (d) CaCO₃ → CaO + CO₂
Answer:
(a) Combination, redox (C: 0 to +4, O: 0 to −2).
(b) Decomposition, redox (Hg: +2 to 0, O: −2 to 0).
(c) Displacement, redox (Zn: 0 to +2, Cu: +2 to 0).
(d) Decomposition, not redox: no oxidation number changes.
In simple words: Always check oxidation numbers; the type of reaction alone does not make it redox.

Teacher's Note:
1. Decomposition and combination are redox only when an element is involved.
2. Write the oxidation number above every atom.

Question 3. Predict all the possible oxidation states of chlorine and give one compound or species for each.
Answer:
Possible states: −1, 0, +1, +3, +5, +7.
−1: HCl; 0: Cl₂; +1: HOCl; +3: HClO₂; +5: HClO₃; +7: HClO₄.
Chlorine has seven valence electrons, so the lowest state is −1 and the highest is +7.
In simple words: Chlorine ranges from −1 to +7, with all the odd values in between.

Teacher's Note:
1. Highest state = group number = 7.
2. Lowest state = 7 − 8 = −1.

Question 4. Calculate the oxidation number of sulphur in H₂SO₄, H₂SO₃, H₂S₂O₈ and H₂S₂O₇.
Answer:
H₂SO₄: 2 + x − 8 = 0, so +6.
H₂SO₃: 2 + x − 6 = 0, so +4.
H₂S₂O₈ has a peroxide link (6 O at −2 and 2 O at −1): 2 + 2x − 12 − 2 = 0, so +6.
H₂S₂O₇: 2 + 2x − 14 = 0, so +6.
In simple words: Only the peroxide oxygens in H₂S₂O₈ are −1; sulphur is still +6.

Teacher's Note:
1. If you used −2 for all eight O in H₂S₂O₈ you would get +7, which is impossible for sulphur.
2. Always look for an O–O bond.

Question 5. Find the oxidation number of every atom in peroxodisulphuric acid, HO–SO₂–O–O–SO₂–OH (H₂S₂O₈).
Answer:
Hydrogen: +1 each.
Sulphur: +6 each.
The six oxygens in the SO₂, OH and S–O positions: −2 each.
The two oxygens joined to each other (peroxide link): −1 each.
Check: 2(+1) + 2(+6) + 6(−2) + 2(−1) = 0.
In simple words: Six oxygens are −2 and the two in the O–O bond are −1.

Teacher's Note:
1. Draw the structure and mark the O–O bond first.
2. Always check that the total is zero.

Question 6. Account for the following: (a) sulphur shows variable oxidation states; (b) fluorine shows only −1; (c) oxygen cannot exhibit a valency of more than 2.
Answer:
(a) Sulphur has vacant d orbitals in its valence shell, so it can use more electrons in bonding and show states such as −2, +4 and +6.
(b) Fluorine is the most electronegative element, so it always attracts the shared pair and is −1 in its compounds.
(c) Oxygen is small and has no vacant d orbitals in its second shell, so it cannot expand its valence beyond 2.
In simple words: Sulphur can expand its octet; fluorine and oxygen cannot.

Teacher's Note:
1. Quote the reason in one line for each part.
2. Fluorine has no positive state in its compounds.

Question 7. Define a redox couple. Identify the couples in Zn + Cu²⁺ → Zn²⁺ + Cu and write the cell in short notation.
Answer:
A redox couple is the oxidised and reduced forms of a substance taking part in an oxidation or reduction half-reaction, written Ox/Red.
The couples are Zn²⁺/Zn and Cu²⁺/Cu.
Cell notation: Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s), the anode on the left and the cathode on the right.
In simple words: Each half-reaction has its own pair: the ion and the metal.

Teacher's Note:
1. | is a phase boundary and || is the salt bridge.
2. Write the oxidised form first in the couple.

Question 8. For the Daniell cell, name the anode and cathode, write the half-reactions and give the direction of electron flow. Calculate E°cell if E°(Zn²⁺/Zn) = −0.76 V and E°(Cu²⁺/Cu) = +0.34 V.
Answer:
Anode (zinc): Zn → Zn²⁺ + 2e⁻.
Cathode (copper): Cu²⁺ + 2e⁻ → Cu.
Electrons flow in the external wire from the zinc electrode to the copper electrode.
E°cell = E°cathode − E°anode = 0.34 − (−0.76) = +1.10 V.
In simple words: Zinc loses electrons and copper gains them, so the cell gives 1.10 V under standard conditions.

Teacher's Note:
1. The positive E°cell shows the reaction is spontaneous.
2. Reduction takes place at the cathode and oxidation at the anode.

Question 9. Explain the role of the salt bridge in a Daniell cell.
Answer:
(a) It completes the electric circuit by allowing ions to migrate between the two half-cells.
(b) It maintains the electrical neutrality of both solutions: anions move towards the anode compartment and cations towards the cathode compartment.
Without it, charge would build up and the cell would stop working quickly.
In simple words: The salt bridge lets ions carry the current inside the cell so that the circuit stays complete.

Teacher's Note:
1. The electrolyte used (KCl or KNO₃) must not react with the ions in either half-cell.
2. Electrons travel in the wire; ions travel through the bridge.

Question 10. What is a standard hydrogen electrode (SHE)? Why is it used?
Answer:
It consists of a platinum foil coated with platinum black, dipping partly in a solution containing H⁺ ions at 1 mol per litre, with hydrogen gas at 1 bar bubbled over it.
The half-reaction is 2H⁺(aq) + 2e⁻ ⇌ H₂(g), and its standard potential is taken as 0.00 V at all temperatures.
The potential of any other electrode cannot be measured alone, so it is measured by joining it to the SHE; the reading is the potential of the other half-cell.
In simple words: The SHE is the zero point on the electrode-potential scale.

Teacher's Note:
1. Mention all the standard conditions: 1 mol per litre, 1 bar and 298 K.
2. Platinum black provides a large surface area for the reaction.

Question 11. Arrange the following in increasing order of reducing power: Cu, Ag, Au, Zn, Fe, Al, Na, Mg, Hg, Ca, K. Which is the strongest and which is the weakest reducing agent?
Answer:
Increasing reducing power: Au < Hg < Ag < Cu < Fe < Zn < Al < Mg < Na < Ca < K.
Strongest reducing agent: K. Weakest: Au.
A metal that is a stronger reducing agent will displace a weaker one from its salt solution.
In simple words: The more easily a metal loses electrons, the stronger it is as a reducing agent.

Teacher's Note:
1. Hydrogen (SHE, 0.00 V) lies between Cu and Fe in this series.
2. Metals above hydrogen displace hydrogen from dilute acids.

Question 12. Is it possible to store (a) a copper sulphate solution in a zinc vessel, (b) a zinc sulphate solution in a copper vessel? Give reasons.
Answer:
(a) No. Zinc is a stronger reducing agent than copper (E° −0.76 V against +0.34 V), so Zn + Cu²⁺ → Zn²⁺ + Cu takes place and the vessel is eaten away.
(b) Yes. Copper cannot reduce Zn²⁺, so no reaction occurs.
In simple words: A metal vessel is safe only if its metal is less reactive than the metal ion in the solution.

Teacher's Note:
1. Compare the positions of the two metals in the series.
2. The vessel must not be able to reduce the ions in the solution.

Question 13. Is the rusting of iron an electrochemical phenomenon? Write the reactions and name two ways of preventing it.
Answer:
Yes. A small electrochemical cell forms on the damp iron surface.
Anode: Fe(s) → Fe²⁺(aq) + 2e⁻.
Cathode: O₂(g) + 4H⁺(aq) + 4e⁻ → 2H₂O(l).
Overall: 2Fe²⁺(aq) + 2H₂O(l) + 1/2 O₂(g) → Fe₂O₃(s) + 4H⁺(aq).
Prevention: (i) galvanisation, i.e. coating with the more reactive zinc, and (ii) painting or greasing to keep out air and moisture.
In simple words: Iron loses electrons at one spot and oxygen gains them at another, which makes it a tiny cell.

Teacher's Note:
1. Rust is hydrated iron(III) oxide.
2. Zinc coating still protects the iron even where it is scratched, because zinc is oxidised in preference to iron.

Question 14. Balance by the ion-electron method in acidic medium: MnO₄⁻ + I⁻ → Mn²⁺ + I₂
Answer:
Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O (×2)
Oxidation: 2I⁻ → I₂ + 2e⁻ (×5)
Adding: 2MnO₄⁻ + 10I⁻ + 16H⁺ → 2Mn²⁺ + 5I₂ + 8H₂O
Check: charge is −2 − 10 + 16 = +4 on the left and +4 on the right.
In simple words: Make the electrons equal at 10 and add the two half-reactions.

Teacher's Note:
1. Balance O with H₂O, then H with H⁺, then charge with electrons.
2. The multipliers 2 and 5 come from the lowest common multiple of 5 and 2.