The electrolysis of other molten substances
4 marks · Normal| Student | Cathode equation | Anode equation |
|---|---|---|
| A | Ca^{2+} + 2e^- \to Ca | 2Br^- \to Br_2 + 2e^- |
| B | Ca^{2+} \to Ca + 2e^- | Br^- \to Br_2 + 2e^- |
| C | Ca^{2+} + e^- \to Ca | 2Br^- \to Br_2 + 2e^- |
| D | Ca^{2+} + 2e^- \to Ca | 2Br^- \to Br_2 + e^- |
Use the table above. For each equation below, state which students wrote it correctly. For each incorrect equation, explain the mistake.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
Show answer
Common mistakes: Omitting electrons; Incorrect numbers of electrons
Show mark scheme and worked solution
Marking points
B1 · 1 mark · Identifies student(s) with correct cathode half-equation (Student A and D only)
B2 · 1 mark · Identifies student(s) with correct anode half-equation (Student A only)
E1 · 1 mark · Explains at least one cathode equation error (e.g., B: electrons on wrong side, not balanced for charge; C: wrong number of electrons)
E2 · 1 mark · Explains at least one anode equation error (e.g., B: not balanced for atoms; D: wrong number of electrons)
Worked solution
Correct cathode equation: Students A and D.
Correct anode equation: Only Student A.
B (cathode): No electrons on the left side, so charges do not balance.
C (cathode): Only one electron, but Ca^{2+} needs to gain 2 electrons to become Ca.
B (anode): Only one Br^- ion used, but Br2 molecule needs two Br atoms; atoms are not balanced.
D (anode): Only one electron given off, but 2Br^- ions collectively lose 2 electrons when forming Br2.