WAEC Chemistry 2012 Theory — Question 5
Question 5 of 6 from the West African Examinations Council (WAEC) Chemistry 2012 Theory paper, with the correct answer and a full explanation.
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5. (Questions 5 and 6 are not for Nigeria in the original paper and are omitted here.) 7(a)(i) Define standard electrode potential. (ii) State two factors that affect the value of standard electrode potential. (iii) Give two uses of the values of standard electrode potential. (iv) Draw and label a diagram for an electrochemical cell made up of Cu²⁺/Cu; E⁰=+0.34V; Zn²⁺/Zn; E⁰=−0.76V. (v) Calculate the e.m.f of the cell in 7(a)(iv) above. (b)(i) In terms of electron transfer, define: I. Oxidation; II. Oxidizing agent. (ii) Balance the following redox reaction: MnO4⁻ + I⁻ + H⁺ → I2 + Mn²⁺. (c) Classify each of the following oxides as basic, amphoteric, acidic or neutral: (i) Carbon (II) Oxide; (ii) Sulphur (IV) oxide; (iii) Aluminium oxide; (iv) Lithium oxide. (d) What is hydrogen bonding?
Model answer
7(a)(i) The standard electrode potential is the potential difference between the electrode and the hydrogen electrode under standard conditions of 298K temperature, 1moldm⁻³ concentration and 1 atm pressure. (ii) Temperature and concentration. (iii) To calculate the emf of an electrochemical cell; to determine the feasibility of a reaction. (iv) The diagram shows a Daniell-type cell: a zinc electrode in ZnSO4 solution connected via a salt bridge to a copper electrode in CuSO4 solution, with a voltmeter (V) connected across the two metal electrodes. (v) The shorthand notation for the cell is Zn/Zn²⁺//Cu²⁺/Cu. E°cell = E°red(Right) − E°red(Left), where E°red(Right) is the standard reduction potential of the compartment on the right and E°red(Left) is that of the compartment on the left. E°cell = 0.34V − (−0.76V) = +1.10V. (b)(i) I. Oxidation is defined as loss of electron(s). II. An oxidizing agent is the substance that gains electron(s). (ii) Reduction half equation: MnO4⁻ → Mn²⁺. Adding 4 moles of H2O to the right and balancing with hydrogen ions: MnO4⁻+8H⁺→Mn²⁺+4H2O. Oxidation half equation: I⁻→I2 i.e. 2I⁻→I2+2e⁻. To make electrons equal, the oxidation half equation is multiplied by 5 while the reduction half equation is multiplied by 2: 10I⁻→5I2+10e⁻; 2MnO4⁻+16H⁺+10e⁻→2Mn²⁺+8H2O. Adding the two and eliminating electrons: 2MnO4⁻+16H⁺+10I⁻→2Mn²⁺+5I2+8H2O. (c)(i) Carbon (II) oxide is neutral. (ii) Sulphur (IV) oxide is acidic. (iii) Aluminium oxide is amphoteric. (iv) Lithium oxide is basic. (d) Hydrogen bonding is an intermolecular force present in molecules where hydrogen is bonded to very small electronegative elements such as nitrogen, oxygen and fluorine.
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