Electrochemistry — A/L Chemistry

How do I practise Electrochemistry?

1 question with a worked solution

Check before you answer: Cell Notation & Standard Potentials: Standard IUPAC cell notation with phase boundaries (|) and salt bridge (||), calculating E°cell = E°cathode - E°anode.

1. For the standard galvanic cell: Zn(s) | Zn²⁺(aq, 1.0 mol dm⁻³) || Cu²⁺(aq, 1.0 mol dm⁻³) | Cu(s), given E°(Zn²⁺/Zn) = -0.76 V and E°(Cu²⁺/Cu) = +0.34 V, what is the standard electromotive force (E°_cell) and the spontaneous oxidation reaction?
   1. A) -1.10 V with reduction of Zn²⁺(aq) + 2e⁻ → Zn(s) at the cathode
   2. B) +1.10 V with oxidation of Zn(s) → Zn²⁺(aq) + 2e⁻ at the anode
   3. C) +0.42 V with reduction of Cu²⁺(aq) + 2e⁻ → Cu(s) at the anode
   4. D) +1.10 V with oxidation of Cu(s) → Cu²⁺(aq) + 2e⁻ at the anode
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   Correct answer: B) +1.10 V with oxidation of Zn(s) → Zn²⁺(aq) + 2e⁻ at the anode

   Zinc has a more negative standard reduction potential (-0.76 V vs +0.34 V for Cu), meaning Zn oxidizes more readily than Cu. Thus, Zn acts as the anode: Zn(s) → Zn²⁺ + 2e⁻ (oxidation). Cu²⁺ acts as the cathode: Cu²⁺ + 2e⁻ → Cu(s) (reduction). E°_cell = E°_cathode - E°_anode = +0.34 V - (-0.76 V) = +1.10 V.

   Show worked solution

   Step 1: Identify cathode and anode from reduction potentials: E°(Cu²⁺/Cu) = +0.34 V (higher potential → reduced at cathode) E°(Zn²⁺/Zn) = -0.76 V (lower potential → oxidized at anode) Step 2: Write half-cell reactions: Anode (oxidation): Zn(s) → Zn²⁺(aq) + 2e⁻ Cathode (reduction): Cu²⁺(aq) + 2e⁻ → Cu(s) Step 3: Calculate cell potential: E°_cell = E°_cathode - E°_anode = (+0.34 V) - (-0.76 V) = +1.10 V. Answer: +1.10 V with oxidation of Zn(s) → Zn²⁺(aq) + 2e⁻ at the anode.
   Modelled on: the A/L Chemistry examination pattern

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Where does Electrochemistry fit in the syllabus?

Grade 13 · Term 2