Purpose: Self-check oxidation numbers, half-reactions, E°cell, and Faraday's law.
Q1
Oxidation number of S in H₂SO₄?
Show solution
2(+1) + S + 4(−2) = 0 → S = +6.
Q2
Oxidation number of Cr in K₂Cr₂O₇.
Show solution
2(+1) + 2Cr + 7(−2) = 0 → 2Cr = +12 → Cr = +6.
Q3
In 2 Na + Cl₂ → 2 NaCl, identify oxidizing agent and reducing agent.
Show solution
Na: 0 → +1 (oxidized) — reducing agent. Cl₂: 0 → −1 (reduced) — oxidizing agent.
Q4
Balance in acidic solution: MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺.
Show solution
Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. Oxidation: Fe²⁺ → Fe³⁺ + e⁻ (×5). Net: MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O.
Q5
Balance in basic solution: Cr(OH)₃ + ClO⁻ → CrO₄²⁻ + Cl⁻.
Show solution
Cr: +3→+6 loses 3e⁻; Cl: +1→−1 gains 2e⁻. Multiply Cr by 2, ClO⁻ by 3. Balanced: 2Cr(OH)₃ + 3ClO⁻ + 4OH⁻ → 2CrO₄²⁻ + 3Cl⁻ + 5H₂O.
Q6
Given E°(Cu²⁺/Cu) = +0.34 V; E°(Zn²⁺/Zn) = −0.76 V. Calculate E°cell for Zn|Zn²⁺‖Cu²⁺|Cu.
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E°cell = E°(cathode) − E°(anode) = (+0.34) − (−0.76) = +1.10 V. Reaction is spontaneous.
Q7
In a galvanic cell, the anode is where:
Show solution
Oxidation occurs at anode (mnemonic: AN OX).
Q8
Write cell notation for Zn|Cu cell described above.
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Zn(s) | Zn²⁺(aq) ‖ Cu²⁺(aq) | Cu(s).
Q9
Calculate moles of electrons in 1.50 A flowing for 30.0 minutes (F = 96485 C/mol).
Show solution
Q = It = 1.50 × 1800 = 2700 C. n = Q/F = 2700/96485 = 0.02798 mol e⁻.
Q10
Mass of Cu deposited from CuSO₄ by 2.00 A for 1.00 h? (Cu²⁺ + 2e⁻ → Cu; M = 63.55)
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Q = 2.00 × 3600 = 7200 C. n(e⁻) = 7200/96485 = 0.0746 mol → n(Cu) = 0.0373 mol → m = 0.0373 × 63.55 = 2.37 g.
Q11
Aluminum is produced via electrolysis: Al³⁺ + 3e⁻ → Al. How many coulombs to produce 1.00 kg Al?
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n(Al) = 1000/26.98 = 37.06 mol; n(e⁻) = 111.2 mol; Q = nF = 111.2 × 96485 = 1.07×10⁷ C.
Q12
A bicycle frame rusts (corrodes) faster in salt-spray. Why?
Show solution
NaCl provides ions that improve electrolyte conductivity, accelerating the electrochemical corrosion process (Fe → Fe²⁺ + 2e⁻, with O₂ + 2H₂O + 4e⁻ → 4OH⁻).