Identify the substances liberated at the anode and at the cathode during the electrolysis of saturated KCl(aq). at the anode at the cathode
- (a)Identify the substances liberated at the anode and at the cathode during the electrolysis of saturated KCl(aq). at the anode at the cathode[1]
- (b)When dilute sulfuric acid is electrolysed, oxygen is liberated at the anode. Dilute sulfuric acid is electrolysed for 15.0 minutes using a current of 0.750A. Calculate the volume of oxygen that is liberated under room conditions. volume of oxygen =[3]
- (c)The halogens chlorine, bromine and iodine differ in their strengths as oxidising agents. These strengths are indicated by the Eº values for these halogens.
- (c)(i)Give the Eº values for chlorine, bromine and iodine acting as oxidising agents.[1]
- (c)(ii)Deduce which of chlorine, bromine and iodine will react with a solution of Sn²⁺(aq) under standard conditions. Explain your answer. Include a relevant equation in your explanation.[3]
- (c)(iii)An excess of chlorine is added to a solution of acidified Mn²⁺(aq) under standard conditions. Give the formula of the product of this reaction that contains manganese.[1]
- (d)An electrochemical cell can be made by connecting an Fe³⁺/Fe²⁺ half-cell to an S2O₈²⁻/SO4²⁻ half-cell under standard conditions.
- (d)(i)Calculate the standard cell potential of this electrochemical cell. Ecell =[1]
- (d)(ii)State the material that should be used as the electrode in each half-cell. in the Fe³⁺/Fe²⁺ half-cell in the S2O₈²⁻/SO4²⁻ half-cell[1]
- (d)(iii)Describe one change to each half-cell that would increase the value of the cell potential. The temperature should remain at 298 K. Fe³⁺/Fe²⁺ half-cell S2O₈²⁻/SO4²⁻ half-cell[1]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2020). The mark or mark type is shown in brackets.
- [1] chlorine AND hydrogen
- [1] 15 × 60 × 0.75 = 675 C
- [1] 675/96 500 = 7.0 × 10-3 moles e¯
- [1] 7.0 × 10-3 × 0.25 gives 1.75 × 10-3 moles O2
- [1] 1.75 x 10-3 × 24000 = 42 (41.969) cm³ O2
- [1] 1.36 1.07 0.54
- [1] all of them
- [1] (all Evalues) greater than 0.15 / Ecell greater than zero
- [1] e.g. Sn2+ + X2 → Sn4+ + 2X
- [1] MnO2
- [1] 1.24 V
- [1] platinum, platinum
- [1] increase [Fe2+] or decrease [Fe3+]
- [1] increase [S2O82] or decrease [SO42-]
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