Question 21M.2.hl.TZ2.3
| Date | May 2021 | Marks available | [Maximum mark: 7] | Reference code | 21M.2.hl.TZ2.3 |
| Level | hl | Paper | 2 | Time zone | TZ2 |
| Command term | Apply, Calculate, Label | Question number | 3 | Adapted from | N/A |
Oxidation and reduction reactions can have a variety of commercial uses.
A student decides to build a voltaic cell consisting of an aluminium electrode, Al (s), a tin electrode, Sn (s), and solutions of aluminium nitrate, Al(NO3)3 (aq) and tin(II) nitrate, Sn(NO3)2 (aq).
Electron flow is represented on the diagram.
Label each line in the diagram using section 25 of the data booklet.
[3]
Al/aluminium «electrode» AND aluminium nitrate/Al(NO3)3/Al3+ on left ✓
Sn/tin «electrode» AND tin«(II)» nitrate/Sn(NO3)2/Sn2+ on right ✓
salt bridge AND voltmeter/V/lightbulb ✓
Award [1] if M1 and M2 are reversed.
Award [1] for two correctly labelled solutions OR two correctly labelled electrodes for M1 and M2.
Accept a specific salt for “salt bridge”.
Accept other circuit components such as ammeter/A, fan, buzzer, resistor/heating element/R/Ω.
Write the equation for the expected overall chemical reaction in (a).
[1]
3Sn2+ (aq) + 2Al (s) → 3Sn (s) + 2Al3+ (aq)
OR
3Sn(NO3)2 (aq) + 2Al (s) → 3Sn (s) + 2Al(NO3)3 (aq) ✓
If half cells are reversed in part-question (a) then the equation must be reversed to award the mark.
Do not penalize equilibrium arrows.
Calculate the cell potential using section 24 of the data booklet.
[1]
«1.66 + (−0.14) = +»1.52 «V» ✓
Calculation must be consistent with equation given in 3 b.
Calculate the Gibbs free energy change, ΔG⦵, in kJ, for the cell, using section 1 of the data booklet.
[2]
«ΔG⦵ = −nFE⦵ = −6 × 9.65 × 104 × 1.52 =» −880080 «J mol−1»
OR
6 «electrons» ✓
«=» −880 «kJ» ✓
Award [1] for “«+»880”.
Award [2] for correct final answer.
