A student investigates the charge (z+) carried by aqueous manganese ions, Mn²+(aq). The electrochemical cell shown is set up for this investigation with the following two half-cells: • a standard copper(II) ion/copper half-cell (E° = +0.340V) • a half-cell made from manganese and 0.500 mol dm⁻³ Mn²+(aq). [Figure 1.1]

1. (a)Label the items P and Q and state the concentration of the copper(II) ion solution in the copper half-cell.[1]
2. (b)(i)Calculate the volume of 0.500 moldm⁻³ Mn²+(aq) needed to prepare 100.0 cm³ of 0.200 mol dm⁻³ Mn²+(aq). volume = .............................................. cm³[1]
3. (b)(ii)Describe how, using a 100cm³ volumetric flask, the student should prepare exactly 100.0 cm³ of 0.200 moldm⁻³ Mn²+(aq) using the volume of 0.500 moldm⁻³ Mn²+(aq) calculated in (b)(i) and standard school or college apparatus.[2]
4. (b)(iii)The cell potential of the electrochemical cell in (a) is measured. The 0.500 mol dm⁻³ Mn²+(aq) is then replaced by the 0.200 moldm⁻³ solution and the cell potential is measured again. This is repeated for other lower concentrations of Mn²+(aq). All measurements are made at 25°C. The results of the experiment are shown in the table. Complete column three of the table, calculating log[Mn²⁺] to two decimal places. Complete column four of the table, calculating E, the electrode potential of each manganese half-cell, to three decimal places, using the equation shown. E(manganese half-cell) = Ecell + 0.340 V | [Mn²⁺] /mol dm⁻³ | cell potential, Ecell/V | log[Mn²⁺] | electrode potential of each manganese half-cell, E/V | | :-------------- | :---------------------- | :-------- | :---------------------------------------------------- | | 5.0 x 10⁻¹ | -1.529 | -0.30 | -1.189 | | 2.0 x 10⁻¹ | -1.541 | | | | 1.0 x 10⁻¹ | -1.550 | | | | 7.5 x 10⁻² | -1.553 | | | | 2.5 x 10⁻² | -1.567 | | | | 8.0 x 10⁻³ | -1.582 | | | | 6.0 x 10⁻³ | -1.590 | | | | 4.0 x 10⁻³ | -1.591 | | | | 3.0 x 10⁻³ | -1.594 | | | | 5.0 x 10⁻⁴ | -1.617 | | |[2]
5. (c)Plot a graph of electrode potential of manganese half-cell (y-axis) against log[Mn²⁺] (x-axis). Use a cross (x) to plot each data point. Draw a line of best fit. [Graph grid for plotting][2]
6. (d)(i)Circle the most anomalous point on your graph.[1]
7. (d)(ii)The student is careful to ensure that all solutions used are at the same temperature in all experiments. Suggest a possible explanation for the position of the anomalous point circled in (d)(i) relative to the line of best fit.[1]
8. (e)Your graph is a plot of E against log[Mn²⁺] and can be analysed using the Nernst equation at 25°C. E = E° + (0.059/Z) log[Mn²⁺] z is the value of the charge carried by the manganese ion E is the electrode potential/V E° is the standard electrode potential/V Use the Nernst equation and your graph to find the standard electrode potential, E°, of the manganese half-cell. E° = .............................................. V[1]
9. (f)(i)Determine the gradient of the graph. State the co-ordinates of both points you used for your calculation. Record the value of the gradient to three significant figures. co-ordinates 1 ..................................................... co-ordinates 2 ..................................................... gradient = ..............................................[2]
10. (f)(ii)Use your answer to (f)(i) and the Nernst equation to calculate the value of z to three significant figures and give the formula of the manganese ion. Your calculation must show the use of the Nernst equation. (If you were unable to calculate an answer to (f)(i) you may use the value 0.0197. This is not the correct value.) z = .............................................. formula of manganese ion = ..............................................[2]
11. (g)Lowering [Mn²⁺] causes the value of the electrode potential of the manganese half-cell to become more negative. Suggest why this happens.[1]

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