Chemistry

1. 1.

   For Mg -> Mg^2+ + 2e^-, identify oxidation or reduction and explain using electrons.

   [3 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Electrons on the product side leave the atom. Electron loss and increasing oxidation state are equivalent descriptions of this change.

   Marking points

   • It is oxidation.
   • Magnesium loses electrons.
   • Its oxidation state increases from zero to +2.

   Examiner tip: Positive ion formation is not reduction merely because electrons appear in the equation.

2. 2.

   Explain the role of a salt bridge in a galvanic cell and why electrons do not travel through it.

   [3 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Without ionic migration, charge buildup opposes further electrode reactions. The bridge supplies an ionic current while the wire supplies an electronic current.

   Marking points

   • Ions move between bridge and half-cells.
   • This maintains charge balance and completes the internal circuit.
   • Electrons flow through the external metallic circuit instead.

   Examiner tip: Avoid saying the bridge transfers electrons or mixes the solutions completely.

3. 3.

   Balance MnO4^- -> Mn^2+ as a reduction half-equation in acidic solution.

   [3 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Balance atoms before charge. The left side without electrons has charge +7 and the right +2, so five electrons must be added on the left.

   Marking points

   • Add 4H2O to the product side to balance oxygen.
   • Add 8H+ to the reactant side to balance hydrogen.
   • MnO4^- + 8H+ + 5e^- -> Mn^2+ + 4H2O.

   Examiner tip: Do not add OH- in the acidic balancing convention.

4. 4.

   Standard reduction potentials are Cu^2+/Cu = +0.34 V and Zn^2+/Zn = -0.76 V. Calculate the standard cell potential and write the spontaneous overall reaction.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Use the more positive reduction potential as the cathode. Subtract the anode's tabulated reduction potential rather than adding both tabulated values.

   Marking points

   • Copper ions are reduced; zinc is oxidised.
   • Ecell = 0.34 - (-0.76) = 1.10 V.
   • Zn + Cu^2+ -> Zn^2+ + Cu.

   Examiner tip: Do not multiply electrode potentials when balancing electron numbers.

5. 5.

   A current of 1.93 A flows for 1000 s through Cu^2+ solution. Assuming 100% current efficiency, calculate deposited copper mass using F = 96500 C mol^-1 and M(Cu) = 63.5 g mol^-1.

   [4 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Apply Q = It, then n(e) = Q/F. The cathode equation Cu^2+ + 2e^- -> Cu supplies the factor of two before converting metal moles to mass.

   Marking points

   • Charge = 1930 C.
   • Electron amount = 0.0200 mol.
   • Copper amount = 0.0100 mol because two electrons reduce each ion.
   • Copper mass = 0.635 g.

   Examiner tip: Faraday's constant converts charge to electron moles, not directly to metal moles.

6. 6.

   A positive standard cell potential predicts feasibility, but an iron object can corrode slowly in dry air. Explain why standard potential alone cannot predict its real corrosion rate.

   [4 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. A redox driving force is only one requirement. Sustained corrosion also needs coupled electrode processes and charge transport; removing moisture can obstruct the ionic part of that circuit.

   Marking points

   • Standard potentials describe thermodynamic tendency under standard conditions.
   • Real concentrations and surface conditions differ.
   • Dry conditions limit an electrolyte path for ionic charge transfer.
   • Activation barriers or surface films can limit reaction rate despite feasible oxidation.

   Examiner tip: Separate thermodynamic feasibility from kinetic and transport limitations.

Original, unofficial practice questions from SubjectScout — not official past-paper questions.