(a) The diagram shows the apparatus used to measure the standard electrode potential, E°, of Fe³⁺(aq)/Fe²⁺(aq). [Figure showing apparatus for standard electrode potential measurement, with labels A-F]
- (a)(i)Identify what the letters A to F represent. A B C D E F[3]
- (a)(ii)Label the diagram to show • which is the positive electrode, • the direction of electron flow in the external circuit. Use the Data Booklet to help you.[1]
- (b)In another experiment, an Fe³⁺(aq)/Fe²⁺(aq) half-cell was connected to a Cu²⁺(aq)/Cu(s) half-cell. Determine the standard cell potential, E°cell, when these two half-cells are connected by a wire and the circuit is completed. Use the Data Booklet to help you. E°cell = V [1][1]
- (c)(i)The E° of Ni²⁺(aq)/Ni(s) is −0.25V. State and explain how the electrode potential changes if the concentration of Ni²⁺(aq) is decreased.[1]
- (c)(ii)The E° of Cr³⁺(aq)/Cr²⁺(aq) is -0.41 V. Calculate the electrode potential when [Cr³⁺(aq)] is 0.60 mol dm⁻³ and [Cr²⁺(aq)] is 0.15 mol dm⁻³. Use the Nernst equation. E = E° + 0.059/z log [oxidised species]/[reduced species] E = V [2][2]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (Feb/Mar 2017). The mark or mark type is shown in brackets.
- [3] A - H2, 1 atm; B - platinum; C - 1 mol dm⁻³ H⁺ / HCl etc.; D - salt bridge / KNO3 etc.; E - platinum; F - 1 mol dm⁻³ Fe²⁺ AND 1 mol dm⁻³ Fe³⁺
- [1] positive electrode is (Pt) on RHS AND electrons flow clockwise
- [1] cell potential is 0.77 – 0.34 =(+) 0.43 (V)
- [1] electrode potential would become more negative as equilibrium shifts to left / explanation in terms of the Nernst equation
- [1] E = −0.41 + (0.059/1)log(0.60/0.15) = -0.41 + 0.059 log 4.0
- [1] E = -0.37 (V)
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