Zadanie 5a, Paper 2, listopad 2025, TZ1: Chemia IB HL

Zadania w arkuszu

Informacja do zadań 5a-5c.

Compounds of transition elements are widely used as electrode materials in secondary (rechargeable) cells, because these elements can exist in more than one oxidation state. Transition elements are also deposited as thin protective or decorative layers by electroplating.

(i) In the nickel–metal hydride (NiMH) rechargeable battery the negative electrode is a metal alloy, M, that absorbs hydrogen. The two electrode reactions during discharge are shown below.

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$\ce{MH(s) + OH-(aq) <=> M(s) + H2O(l) + e-}$

$\ce{NiO(OH)(s) + H2O(l) + e- <=> Ni(OH)2(s) + OH-(aq)}$

Deduce the overall cell equation.

(ii) Simple voltaic cells make use of the difference in standard electrode potential between two half-cells. Shown below is an incomplete set-up for measuring the cell potential between $\ce{Fe^2+(aq)}$ / $\ce{Fe}$ and $\ce{Cu^2+(aq)}$ / $\ce{Cu}$ half-cells.

incomplete voltaic cell, two beakers each containing a solution and an electrode, labelled anode and cathode, with blank label lines pointing to the electrodes and the solutions

Draw in the missing components, and add full labels, so that the diagram shows how this cell potential could actually be measured.

(iii) For this cell, calculate the standard cell potential, $E^\ominus_\mathrm{cell}$ . Use section 19 of the data booklet.

Sprawdzane umiejętności

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Odpowiedź z klucza

(i) $\ce{MH(s) + NiO(OH)(s) -> M(s) + Ni(OH)2(s)}$ ✔

(ii) salt bridge connecting the two solutions ✔

voltmeter connected to the two electrodes by wires ✔

electrodes labelled Fe «anode» AND Cu «cathode» AND solutions labelled $\ce{Fe^2+(aq)}$ AND $\ce{Cu^2+(aq)}$

(iii) $E^\ominus_\mathrm{cell}$ = «+0.34 − (−0.45) =» +0.79 «V» ✔

Schemat punktowania

(i) [1] for the overall equation with water and hydroxide ions cancelled; accept an equilibrium arrow. (ii) M1: salt bridge in contact with both solutions; M2: voltmeter and wires connecting both electrodes; M3: electrodes and ions in solution correctly labelled. Ignore the direction of electron flow and standard conditions. (iii) [1] for +0.79 V.

Komentarz

Iron has the more negative electrode potential, so it is oxidized at the anode; the cell potential is the difference between the two standard electrode potentials, not their sum.

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