IBChemistry: HLDPExam Questions19. Redox Processes (HL only)19.1 Electrochemical Cells

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First teaching 2014

Last exams 2024

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Electrochemical Cells (DP IB Chemistry: HL)

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19.1 Electrochemical Cells

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19.1 Electrochemical Cells

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1
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Consider these standard electrode potentials.

Fe2+ (aq) + 2e- rightwards harpoon over leftwards harpoonFe (s) EΘ = -0.45 V

Cu2+ (aq) + 2e- rightwards harpoon over leftwards harpoonCu (s) EΘ = +0.15 V

Which is the correct working to determine EΘcell?

  • EΘcell = 0.15 - (-0.45)

  • EΘcell = 0.15 + (-0.45)

  • EΘcell = (-0.45) - 0.15

  • EΘcell = 0.15 x (-0.45)

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2
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Which of the following is not a condition for the standard hydrogen electrode (SHE)?

  • 1.00 mol dm-3 HCl 

  • Hydrogen gas with a pressure of 100 Pa

  • Temperature of 298 K 

  • Platinum electrodes 

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3
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Which combination would electroplate an object with silver?

bf04f6dc-ae51-4609-b897-777b9a2ab6f6

  X Y Z
A. Object to be plated Silver Silver chloride
B. Silver Object to be plated Hydrochloric acid 
C. Object to be plated Silver Water
D. Silver Object to be plated Silver nitrate

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    4
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    What are the products for the electrolysis of concentrated sodium chloride solution using inert electrodes?

      Anode Cathode
    A. O2 (g) H2 (g)
    B. H2 (g) O2 (g)
    C. Cl2 (g) H2 (g)
    D. H2 (g) Cl2 (g)

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      5
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      A voltaic cell is made by connecting two half-cells represented by the half-equations below.

      Al3+ (aq) + 3e- rightwards harpoon over leftwards harpoonAl (s) EΘ = -1.66 V

      Sn2+ (aq) + 2e- rightwards harpoon over leftwards harpoonSn (s) EΘ = +0.14 V

      Which statement is correct about this voltaic cell?

      • The cell representation is Al (s) l Al3+ (aq) II Sn2+ (aq) I Sn (s)

      • The Al3+ (aq) / Al (s) electrode is the cathode 

      • The cell representation is Al3+ (aq) l Al (s) II Sn (s) I Sn2+ (s)

      • The Sn2+ (aq) / Sn (s) electrode is the anode 

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      11 mark

      Use the following electrode potentials to answer the question.

      Sn2+ (aq) + 2e- ⇌ Sn (s)      Eθ = -0.14 V

      Fe3+ (aq) + e- ⇌  Fe2+ (aq)  Eθ = +0.77 V

      What will be the EMF, in V, when the following voltaic cell is connected?

      Sn (s) + 2Fe3+ (aq) → 2Fe2+ (aq) + Sn2+ (aq)

      • -0.91

      • +0.63

      • +1.68

      • +0.91

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      21 mark

      Which of the following reactions could take place at the positive electrode (cathode) in a voltaic cell? 

      1.   Cu2+ (aq) to Cu (s)
      2.   Br2 (g) to Br- (aq)
      3.   Co3+ (aq) to Co2+(aq)

      • I and II only

      • I and III only

      • II and III only

      • I, II and III

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      31 mark

      What is true when aqueous copper(II) sulfate is electrolysed using platinum electrodes?

      • H2 and O2 are produced in a 2:1 mole ratio

      • Cu and O2 are produced in a 2:1 mole ratio

      • H2 and O2 are produced in a 1:1 mole ratio

      • Cu and O2 are produced in a 1:1 mole ratio

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      41 mark

      Use the following electrode potentials to answer the question. 

      Zn2+ (aq) + 2e- ⇌ Zn (s)      Eθ = -0.76 V

      Cl2 (aq) + 2e- ⇌  2Cl- (aq)   Eθ = +1.36 V

      Mg2+ (aq) + 2e- ⇌ Mg (s)     Eθ = -2.37 V 

      Predict what happens when some powdered zinc is added to aqueous magnesium chloride?

      • There is no reaction observed

      • Bubbles of chlorine gas will be seen

      • Magnesium metal will be produced

      • Zinc chloride will be produced

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      51 mark

      Which of the following electrolytic cells would give the greatest mass of metal at the cathode?

       

      Current

      Time

      Solution

      A.

      1.5

      250

      1.0 mol dm-3 AgNO3 (aq)

      B.

      1.0

      750

      1.0 mol dm-3 CuSO4 (aq)

      C.

      2.0

      250

      1.0 mol dm-3 AgNO3 (aq)

      D.

      1.0

      500

      1.0 mol dm-3 CuSO4 (aq)

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        1
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        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

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          2
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          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

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            3
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            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

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            4
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            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

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            5
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            The oxidation of iron is a spontaneous process described by the overall equation:

            Fe (s) + bevelled 1 halfO2 (g) + H2O (l) → Fe(OH)2 (s)       

            (ΔGθ = -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 styleO2 (g) + H2O (l) + 2e- → 2OH- (s)

            (ΔGθ = -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

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