Topic 19 – Redox Processes (AHL)

WITHIN TOPIC QUESTIONS

Topic 19 – Redox processes (AHL) Quick questions Page 427 The sludge might contain other metals, such as silver, gold and platinum; these elements can be extracted from the sludge by chemical reagents, reduced and isolated as individual metals; therefore, the price of the sludge is often higher than that of copper; Page 429 Pure water is a very weak conductor of electricity, so an electrolyte is needed for the electrolysis of

water; dilute H2SO4 is a strong acid, so it fully dissociates in aqueous solutions and produces enough ions (carriers of electrical charge) for the electrolysis to proceed; in addition, the anion of sulfuric acid, 2− SO4 (aq), does not undergo any chemical changes during the electrolysis, so it does not need to be replaced;

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End of topic questions (page 434) 1. D; these three conditions constitute the definition of a standard cell; 2. A; zinc is less active than magnesium, so zinc can neither substitute magnesium (answer C is incorrect) or from a salt (answer D is incorrect) under these conditions; to displace chlorine, a more active halogen (not metal) is needed, so answer B is also incorrect; please note, however, that the solution of magnesium chloride is slightly acidic (due to hydrolysis), so zinc might slowly react with water to produce hydrogen gas;

3. D; aluminium is more active than nickel, so Al(s) will be oxidized to Al3+(aq) while Ni2+(aq) will be reduced to Ni(s); the standard cell potential will be equal to the difference of the standard half-cell potentials: Eϴ = −0.23 − (−1.66) = 1.43 V 4. B; in both cases, elemental bromine will be produced at the anode; in the case of molten magnesium bromide, only magnesium metal can be produced at the cathode; in contrast, the electrolysis of aqueous magnesium bromide will produce hydrogen gas at the cathode;

5. C; the reduction of Ag+(aq) to Ag(s) can take place only at the negative electrode (cathode); 6. B; to answer this question, we need to look at the half-equations (the states are omitted):

Na + + e− → Na 2Br − Br 2e− → 2 + 2 Mg + + 2e− → Mg 2Cl − Cl 2e− → 2 + since the same quantity of electricity was used, the amounts of products will be inversely proportional to the numbers of electrons involved in the half-equations: n(Na) 2n(Br ) 2n(Mg) 2n(Cl ) = 2 = = 2 z F m z F m 7. A; according to Faraday’s Law, Q _× × , where Q I t; therefore, I t _× × and so = M = × × = M _M I t 2 1 m = × × ; for Cu +(aq)/Cu(s), M = 63.55 g mol− and z = 2; finally,t = 60 × 60 = 3600 s, z × F 1 I = 1 A, and F = 96 500 C mol− 8. the voltaic cell can be represented as follows: voltmeter

e- anode cathode ( ) V - (+) Co(s) anions Ag(s)

salt bridge

2 Co +(aq) Ag+(aq)

(i) cathode (positive electrode): Ag+(aq) + e− → Ag(s) 2 anode (negative electrode): Co(s) → Co +(aq) + 2e− 2 (ii) Co(s) + 2Ag+(aq) → Co +(aq) + 2Ag(s) 2 (iii) Co(s) | Co +(aq) || Ag+(aq) | Ag(s) (iv) electrons move from the cobalt electrode (anode) to the silver electrode (cathode) through the external circuit; the Co2+(aq) ions are formed from Co(s) at the anode and move into the aqueous solution in the left hand side beaker; the Ag+(aq) ions in the aqueous solution

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in the right hand side beaker move towards the silver cathode and form Ag(s); the counter- ions (anions) move through the salt bridge from the right hand side beaker to the left hand side beaker;

(v) according to the IB Data Booklet, the Eϴ for Ag+(aq)/Ag(s) is +0.80 V; therefore, the cell potential is +0.80 − (−0.28) = 1.08 V (vi) ΔGϴ = −nFEϴ = −2 × 96 500 × 1.08 ≈ −208 000 J = −208 kJ 9. The electrolytic cell can be represented as follows: power source anode cathode + - e- ( ) + (-) Pt(s) Pt(s)

H2(g) KI(aq)

H+(aq) I2(aq) I-(aq)

Note 1: depending on the experimental conditions, elemental iodine can be produced as I2(aq) or

I2(s), or both; Note 2: elemental iodine is readily soluble in concentrated KI(aq) due to the formation of triiodide anions: I−(aq) I (aq) I −(aq); this reaction is discussed in Options B.4 and C.8; it is not required + 2 ⇋ 3 for AHL;

(i) cathode (negative electrode): 2H O(l) 2e− H (g) 2OH−(aq) 2 + → 2 + anode (positive electrode): 2I−(aq) I (aq) 2e− → 2 + or

3I −(aq) I −(aq) 2e− → 3 + overall cell reaction: 2H O(l) 2I−(aq) H (g) I (aq) 2OH−(aq) 2 + → 2 + 2 + or

2H O(l) 3I−(aq) H (g) I −(aq) 2OH−(aq) 2 + → 2 + 3 + Note: the equations involving triiodide ions are not required for AHL; (ii) any inert conducting material, such as platinum or graphite;

(iii) electrons move from the anode to the cathode through the external circuit; the I−(aq) ions move to the anode; the H+(aq) ions (produced from water) move to the cathode; (iv) at the anode, the initially colourless solution will turn first yellow, then orange, and finally − brown due to the formation of I2(aq) and I3 (aq) ions; at high electric currents, the formation

of grey-coloured precipitate of I2(s) can be observed; at the cathode, bubbles of hydrogen gas will be formed; 10. a) power source anode cathode + - e- ( ) + (-) Pt(s) Pt(s)

O2(g) H2(g)

H2SO4(aq)

H+(aq) OH-(aq)

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b) cathode (negative electrode): 2H+(aq) 2e− H (g) + → 2 anode (positive electrode): 2H O(l) O (g) 4H+(aq) 4e− 2 → 2 + + overall cell reaction: 2H O(l) 2H (g) O (g) 2 → 2 + 2 c) electrons move from the anode to the cathode through the external circuit; the H+(aq) ions move to the cathode; the OH−(aq) ions (produced from water) move to the anode; _V d) according to Faraday’s Law, Q = z × F × n, where Q = I × t and n = ; therefore, VM z F V I t VM I t _× × ; and so V _× × ; according to the IB Data Booklet, the molar volume of × = V = z F M × ideal gas at STP (273 K and 100 kPa) is 22.7 dm3 mol−1, so at SATP (298 K and 100 kPa), 22.7 298 V __× 24.8 dm3 mol−1; finally,F 96 500 Q mol−1 and 1 Q 1 A 1 s, so the M = 273 ≈ = = × 1 4 electrolysis duration must be converted to seconds: 5.00 h × 3600 s h− = 1.80 × 10 s; for 2.35 1.80 104 24.8 H (g), z 2 and so V(H ) ___× × × 5.44 dm3; according to the overall cell 2 = 2 = ≈ 2 × 96500 reaction in (b), the volume of oxygen will be twice as low (2.72 dm3);

2 11. tin cathode (negative electrode): Cu +(aq) + 2e− → Cu(s) 2 copper anode (positive electrode): Cu(s) → Cu +(aq) + 2e− Observations: the tin cathode changes colour from silvery to red, becomes thicker, and its mass gradually increases (due to copper deposition); the copper anode becomes thinner and its mass gradually decreases (due to copper passing into the solution); the blue colour of the solution does not change (as the amount of copper ions consumed at the cathode is equal to the amount of copper ions produced at the anode); the temperature of the solution may rise (as some electrical energy will be converted to heat due to non-zero electrical resistance of the electrolyte); 12. In the first cell, magnesium metal will be produced at the cathode (negative electrode):

2 Mg + + 2e− → Mg(l)

2+ Note: the state of the Mg ion in molten MgCl2 is difficult to define, so it is usually omitted. In the second cell, the following processes will take place:

cathode (negative electrode): 2H O(l) 2e− H (g) 2OH−(aq) 2 + → 2 + anode (positive electrode): 4OH−(aq) O (g) 2H O(l) 4e− → 2 + 2 + overall cell reaction: 2H O(l) 2H (g) O (g) 2 → 2 + 2 2 m(H ) 4 m(O ) __2 × m(Mg) _× 2 _× 2 According to Faraday’s Law, = = , where the coefficients “2” Ar(Mg) Mr(H2) Mr(O2) and “4” are the numbers of electrons involved in respective half-equations; therefore, 12.16 2.02 2 12.16 32.00 2 m(H ) __× × 1.01 g and m(O ) __× × 8.00g. 2 = ≈ 2 = ≈ 24.31 × 2 24.31 × 4

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