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DP IB Chemistry: HL

Topic Questions

Home / IB / Chemistry: HL / DP / Topic Questions / 19. Redox Processes (HL only) / 19.1 Electrochemical Cells / Multiple Choice: Paper 1


19.1 Electrochemical Cells

Question 1

Marks: 1

Which of the following can be used for a standard hydrogen electrode (SHE)?

  Electrode Electrolyte solution
A. Graphite 1 mol dm-3 H2SO4
B. Graphite 1 mol dm-3 HCl
C. Platinum 0.5 mol dm-3 H2SO4
D. Platinum 0.5 mol dm-3 HCl
    Choose your answer
      

    Question 2

    Marks: 1

    What are the ratios of gases produced at the electrodes in the electrolysis of dilute vs concentrated sodium chloride solution?

      Ratio of gas produced at cathode : anode in the electrolysis of dilute NaCl Ratio of gas produced at cathode : anode in the electrolysis of concentrated NaCl
    A. 1:1 2:1
    B. 1:1 1:2
    C. 1:2 1:1
    D. 2:1 1:1
      Choose your answer
        

      Question 3

      Marks: 1

      Which of the following could be used to electroplate a zinc medal?

      • 1.0 mol dm-3 AgNO3 solution with a silver anode

      • 1.0 mol dm-3 AgNO3 with a silver cathode

      • 1.0 mol dm-3 CuSO4 solution with a copper cathode

      • 1.0 mol dm-3 solution of SnCl2 and a tin anode

      Choose your answer
        
      Key Concepts
      Electroplating

      Question 4

      Marks: 1

      The diagram below shows the set-up of aluminium and silver cells in series:

      silver-and-aluminium-cells-in-series-ib-hl-mcq-19-1-q4

      What calculation shows the loss in mass of the aluminium electrode if the silver electrode gains 0.25 g?

      • Al mass lost = fraction numerator 107.87 space cross times space 3 space cross times 0.25 over denominator 26.98 end fraction

      • Al mass lost = fraction numerator 107.87 space cross times space 0.33 space cross times 0.25 over denominator 26.98 end fraction

      • Al mass lost = fraction numerator 26.98 space cross times 0.33 space cross times 0.25 over denominator 107.87 end fraction

      • Al mass lost = fraction numerator 26.98 space cross times 3 space cross times 0.25 over denominator 107.87 end fraction

      Choose your answer
        

      Question 5

      Marks: 1

      The oxidation of iron is a spontaneous process described by the overall equation:

      Fe (s) + bevelled 1 halfO(g) + H2O (l) → Fe(OH)2 (s)       

      Gθ = -164 kJ mol-1 at 298 K)

      The two half equations for the process are:

      Fe2+ (s) + 2e- → Fe (s)

      begin mathsize 10px style bevelled 1 half end styleO(g) + H2O (l) + 2e- → 2OH- (s)

      Gθ = -nFEθ , F = 9.65 x 104  C mol-1)

      Which is the correct calculation to work out Eθ in V?

      • Eθ fraction numerator negative 164 over denominator negative 2 space cross times 9.65 end fraction

      • Eθ fraction numerator negative 164 space 000 over denominator negative 2 space cross times 96 space 500 end fraction

      • Eθ fraction numerator negative 164 space 000 over denominator negative 0.5 space cross times space 96 space 500 end fraction

      • Eθ fraction numerator negative 164 over denominator negative 0.5 space cross times 9.65 end fraction

      Choose your answer
        
      Key Concepts
      Free Energy & Eº