Intext Questions
1. How would you determine the standard electrode potential of the system Mg²⁺|Mg?
Answer
Set up a galvanic cell with a magnesium strip dipped in 1 M Mg²⁺ solution (e.g. MgSO₄) as one electrode and the standard hydrogen electrode (Pt, H₂ at 1 bar, 1 M H⁺) as the other, joined by a salt bridge: Mg(s) | Mg²⁺(1 M) || H⁺(1 M) | H₂(1 bar) | Pt(s).
Measure the emf with a voltmeter. The voltmeter shows that Mg is the anode (electrons flow from Mg to SHE). Since E°cell = E°(SHE) − E°(Mg²⁺/Mg) = 0 − E°(Mg²⁺/Mg), the measured emf (≈ 2.36 V) gives E°(Mg²⁺/Mg) = −2.36 V.
2. Can you store copper sulphate solutions in a zinc pot?
Answer
No. E°(Zn²⁺/Zn) = −0.76 V is lower than E°(Cu²⁺/Cu) = +0.34 V, so zinc is a stronger reducing agent and displaces copper: Zn(s) + Cu²⁺(aq)
⇒ Zn²⁺(aq) + Cu(s). The pot would be corroded and the solution spoiled.
3. Consult the table of standard electrode potentials and suggest three substances that can oxidise ferrous ions under suitable conditions.
Answer
Fe²⁺
⇒ Fe³⁺ + e⁻ has E°(Fe³⁺/Fe²⁺) = 0.77 V. Any species with a higher reduction potential can oxidise Fe²⁺, for example: Cl₂ (1.36 V), Br₂ (1.09 V), acidified K₂Cr₂O₇ (1.33 V), acidified KMnO₄ (1.51 V), H₂O₂ (1.78 V), O₂ in acid (1.23 V). Any three of these.
4. Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.
Answer
H⁺ + e⁻
⇒ ½H₂; [H⁺] = 10⁻¹⁰ M, p(H₂) = 1 bar
E = E° − 0.059 log(1/[H⁺]) = 0 − 0.059 × 10 = −0.59 V
5. Calculate the emf of the cell in which the following reaction takes place: Ni(s) + 2Ag⁺ (0.002 M) → Ni²⁺ (0.160 M) + 2Ag(s). Given that E°cell = 1.05 V.
Answer
n = 2. E = E° − (0.059/2) log([Ni²⁺]/[Ag⁺]²)
= 1.05 − 0.0295 log(0.160/(0.002)²) = 1.05 − 0.0295 log(4 × 10⁴) = 1.05 − 0.0295 × 4.602 = 1.05 − 0.136 = 0.914 V
6. The cell in which the following reaction occurs: 2Fe³⁺(aq) + 2I⁻(aq) → 2Fe²⁺(aq) + I₂(s) has E°cell = 0.236 V at 298 K. Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.
Answer
n = 2. ΔrG° = −nFE° = −2 × 96487 × 0.236 = −45542 J mol⁻¹ = −45.54 kJ mol⁻¹
log K = −ΔrG°/(2.303RT) = 45542/(2.303 × 8.314 × 298) = 7.98
⇒ K = 9.62 × 10⁷
7. Why does the conductivity of a solution decrease with dilution?
Answer
Conductivity is the conductance of 1 cm³ of solution. On dilution, the number of ions present per unit volume decreases, so fewer ions carry current in that volume and conductivity falls (even though molar conductivity increases).
8. Suggest a way to determine the Λ°m value of water.
Answer
Use Kohlrausch's law of independent migration of ions: Λ°m(H₂O) = λ°(H⁺) + λ°(OH⁻) = Λ°m(HCl) + Λ°m(NaOH) − Λ°m(NaCl). The Λ°m values of these strong electrolytes are found by extrapolating Λm versus √c plots to zero concentration.
9. The molar conductivity of 0.025 mol L⁻¹ methanoic acid is 46.1 S cm² mol⁻¹. Calculate its degree of dissociation and dissociation constant. Given λ°(H⁺) = 349.6 S cm² mol⁻¹ and λ°(HCOO⁻) = 54.6 S cm² mol⁻¹.
Answer
Λ°m = 349.6 + 54.6 = 404.2 S cm² mol⁻¹
α = Λm/Λ°m = 46.1/404.2 = 0.114
Ka = cα²/(1 − α) = 0.025 × (0.114)²/(0.886) = 3.67 × 10⁻⁴ mol L⁻¹
10. If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire?
Answer
Q = I × t = 0.5 × 2 × 3600 = 3600 C
Number of electrons = Q/e = 3600/(1.602 × 10⁻¹⁹) = 2.25 × 10²²
11. Suggest a list of metals that are extracted electrolytically.
Answer
Highly reactive metals with very negative electrode potentials: sodium, potassium, lithium, calcium, magnesium and aluminium (by electrolysis of their fused salts/oxides).
12. Consider the reaction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O. What is the quantity of electricity in coulombs needed to reduce 1 mol of Cr₂O₇²⁻?
Answer
6 mol electrons are needed = 6 F = 6 × 96487 C = 5.79 × 10⁵ C
13. Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging.
Answer
During discharge both electrodes get coated with PbSO₄ and H₂SO₄ is used up. On recharging, an external current is passed in the reverse direction and the cell works as an electrolytic cell:
At the negative plate (cathode during charging): PbSO₄(s) + 2e⁻
⇒ Pb(s) + SO₄²⁻(aq)
At the positive plate (anode during charging): PbSO₄(s) + 2H₂O(l)
⇒ PbO₂(s) + SO₄²⁻(aq) + 4H⁺(aq) + 2e⁻
Overall: 2PbSO₄(s) + 2H₂O(l)
⇒ Pb(s) + PbO₂(s) + 2H₂SO₄(aq)
Materials involved: lead sulphate on both plates, water, the regenerated spongy lead (anode), lead dioxide (cathode) and sulphuric acid (electrolyte, whose concentration rises).
14. Suggest two materials other than hydrogen that can be used as fuels in fuel cells.
Answer
Methane and methanol (ethanol can also be used).
15. Explain how rusting of iron is envisaged as setting up of an electrochemical cell.
Answer
At a spot on the iron surface, iron is oxidised and acts as anode: 2Fe(s)
⇒ 2Fe²⁺ + 4e⁻ (E° = −0.44 V for Fe²⁺/Fe). The electrons travel through the metal to another spot that acts as cathode, where oxygen is reduced in the presence of H⁺ (from H₂CO₃ formed by CO₂ in water): O₂(g) + 4H⁺ + 4e⁻
⇒ 2H₂O (E° = 1.23 V). Overall: 2Fe + O₂ + 4H⁺
⇒ 2Fe²⁺ + 2H₂O, E°cell = 1.67 V. The moisture film acts as electrolyte. The Fe²⁺ ions are further oxidised by air to Fe³⁺ and form hydrated ferric oxide, Fe₂O₃·xH₂O (rust).
Exercises
1. Arrange the following metals in the order in which they displace each other from the solution of their salts. Al, Cu, Fe, Mg and Zn.
Answer
A metal with a lower (more negative) E° displaces those with higher E°. E°: Mg −2.36, Al −1.66, Zn −0.76, Fe −0.44, Cu +0.34 V.
Order: Mg > Al > Zn > Fe > Cu (Mg displaces all others; Cu displaces none).
2. Given the standard electrode potentials, K⁺/K = −2.93 V, Ag⁺/Ag = 0.80 V, Hg²⁺/Hg = 0.79 V, Mg²⁺/Mg = −2.37 V, Cr³⁺/Cr = −0.74 V. Arrange these metals in their increasing order of reducing power.
Answer
Lower reduction potential means stronger reducing agent.
Increasing reducing power: Ag < Hg < Cr < Mg < K
3. Depict the galvanic cell in which the reaction Zn(s) + 2Ag⁺(aq) → Zn²⁺(aq) + 2Ag(s) takes place. Further show: (i) Which of the electrode is negatively charged? (ii) The carriers of the current in the cell. (iii) Individual reaction at each electrode.
Answer
Cell: Zn(s) | Zn²⁺(aq) || Ag⁺(aq) | Ag(s)
(i) The zinc electrode (anode) is negatively charged.
(ii) In the external circuit current is carried by electrons (flowing from Zn to Ag); inside the cell and the salt bridge it is carried by ions (Zn²⁺, Ag⁺ and the ions of the salt bridge).
(iii) Anode (oxidation): Zn(s)
⇒ Zn²⁺(aq) + 2e⁻; Cathode (reduction): Ag⁺(aq) + e⁻
⇒ Ag(s) (×2).
4. Calculate the standard cell potentials of galvanic cell in which the following reactions take place: (i) 2Cr(s) + 3Cd²⁺(aq) → 2Cr³⁺(aq) + 3Cd (ii) Fe²⁺(aq) + Ag⁺(aq) → Fe³⁺(aq) + Ag(s). Calculate the ΔrG° and equilibrium constant of the reactions.
Answer
(i) E°(Cd²⁺/Cd) = −0.40 V, E°(Cr³⁺/Cr) = −0.74 V; n = 6
E°cell = −0.40 − (−0.74) = 0.34 V
⇒ ΔrG° = −6 × 96487 × 0.34 = −196.8 kJ mol⁻¹
log K = 196833/(2.303 × 8.314 × 298) = 34.50
⇒ K ≈ 3.2 × 10³⁴
(ii) E°(Ag⁺/Ag) = 0.80 V, E°(Fe³⁺/Fe²⁺) = 0.77 V; n = 1
E°cell = 0.80 − 0.77 = 0.03 V
⇒ ΔrG° = −1 × 96487 × 0.03 = −2.89 kJ mol⁻¹
log K = 2894.6/(2.303 × 8.314 × 298) = 0.507
⇒ K ≈ 3.22
5. Write the Nernst equation and emf of the following cells at 298 K: (i) Mg(s)|Mg²⁺(0.001 M)||Cu²⁺(0.0001 M)|Cu(s) (ii) Fe(s)|Fe²⁺(0.001 M)||H⁺(1 M)|H₂(g)(1 bar)|Pt(s) (iii) Sn(s)|Sn²⁺(0.050 M)||H⁺(0.020 M)|H₂(g)(1 bar)|Pt(s) (iv) Pt(s)|Br⁻(0.010 M)|Br₂(l)||H⁺(0.030 M)|H₂(g)(1 bar)|Pt(s).
Answer
(i) Mg + Cu²⁺
⇒ Mg²⁺ + Cu; n = 2; E° = 0.34 − (−2.36) = 2.70 V
E = E° − (0.059/2) log([Mg²⁺]/[Cu²⁺]) = 2.70 − 0.0295 log(0.001/0.0001) = 2.70 − 0.0295 = 2.67 V
(ii) Fe + 2H⁺
⇒ Fe²⁺ + H₂; n = 2; E° = 0 − (−0.44) = 0.44 V
E = 0.44 − 0.0295 log([Fe²⁺]/[H⁺]²) = 0.44 − 0.0295 log(0.001/1) = 0.44 + 0.0885 = 0.53 V
(iii) Sn + 2H⁺
⇒ Sn²⁺ + H₂; n = 2; E° = 0 − (−0.14) = 0.14 V
E = 0.14 − 0.0295 log(0.050/(0.020)²) = 0.14 − 0.0295 log 125 = 0.14 − 0.062 = 0.078 V
(iv) 2Br⁻ + 2H⁺
⇒ Br₂ + H₂; n = 2; E° = 0 − 1.09 = −1.09 V
E = −1.09 − 0.0295 log(1/([Br⁻]²[H⁺]²)) = −1.09 − 0.0295 log(1/((0.010)² × (0.030)²)) = −1.09 − 0.0295 × 7.05 = −1.09 − 0.208 = −1.298 V
The negative emf shows the cell reaction as written is not spontaneous; the reverse reaction would occur.
6. In the button cells widely used in watches and other devices the following reaction takes place: Zn(s) + Ag₂O(s) + H₂O(l) → Zn²⁺(aq) + 2Ag(s) + 2OH⁻(aq). Determine ΔrG° and E° for the reaction.
Answer
Anode: Zn
⇒ Zn²⁺ + 2e⁻, E°(Zn²⁺/Zn) = −0.76 V. Cathode: Ag₂O + H₂O + 2e⁻
⇒ 2Ag + 2OH⁻, E° = 0.344 V
E°cell = 0.344 − (−0.76) = 1.104 V
ΔrG° = −nFE° = −2 × 96487 × 1.104 = −2.13 × 10⁵ J mol⁻¹ = −213.0 kJ mol⁻¹
7. Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.
Answer
Conductivity (κ) is the reciprocal of resistivity; it is the conductance of a solution of unit volume, i.e. placed between two electrodes of 1 cm² area kept 1 cm apart (unit S cm⁻¹ or S m⁻¹).
Molar conductivity (Λm) is the conductance of the volume of solution containing 1 mol of electrolyte, kept between electrodes of area A at unit distance: Λm = κ/c (Λm in S cm² mol⁻¹ = κ × 1000/M).
Variation: conductivity decreases on dilution because ions per unit volume decrease. Molar conductivity increases on dilution because the total volume containing one mole increases more than κ decreases. For strong electrolytes Λm increases slowly and linearly with √c, following Λm = Λ°m − A√c. For weak electrolytes Λm rises steeply at low concentration because the degree of dissociation increases sharply; Λ°m cannot be found by extrapolation and is obtained from Kohlrausch's law.
8. The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm⁻¹. Calculate its molar conductivity.
Answer
Λm = κ × 1000/M = 0.0248 × 1000/0.20 = 124 S cm² mol⁻¹
9. The resistance of a conductivity cell containing 0.001 M KCl solution at 298 K is 1500 Ω. What is the cell constant if conductivity of 0.001 M KCl solution at 298 K is 0.146 × 10⁻³ S cm⁻¹.
Answer
Cell constant G* = κ × R = 0.146 × 10⁻³ × 1500 = 0.219 cm⁻¹
10. The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below: Concentration/M: 0.001, 0.010, 0.020, 0.050, 0.100; 10² × κ/S m⁻¹: 1.237, 11.85, 23.15, 55.53, 106.74. Calculate Λm for all concentrations and draw a plot between Λm and c½. Find the value of Λ°m.
Answer
Λm = κ/(1000c) in S m² mol⁻¹ when c is in mol L⁻¹ (1 mol L⁻¹ = 1000 mol m⁻³). Converting to S cm² mol⁻¹ (× 10⁴):
c = 0.001 M: Λm = 1.237 × 10⁻²/1 = 1.237 × 10⁻² S m² mol⁻¹ = 123.7 S cm² mol⁻¹; √c = 0.0316
c = 0.010 M: Λm = 11.85 × 10⁻²/10 = 118.5 S cm² mol⁻¹; √c = 0.100
c = 0.020 M: Λm = 23.15 × 10⁻²/20 = 115.8 S cm² mol⁻¹; √c = 0.141
c = 0.050 M: Λm = 55.53 × 10⁻²/50 = 111.1 S cm² mol⁻¹; √c = 0.224
c = 0.100 M: Λm = 106.74 × 10⁻²/100 = 106.7 S cm² mol⁻¹; √c = 0.316
A plot of Λm against √c is nearly a straight line (strong electrolyte). Extending it to √c = 0, the intercept gives Λ°m ≈ 124 S cm² mol⁻¹ (1.24 × 10⁻² S m² mol⁻¹). The slope (−A) is about −60 S cm² mol⁻¹ (mol L⁻¹)⁻½.
11. Conductivity of 0.00241 M acetic acid is 7.896 × 10⁻⁵ S cm⁻¹. Calculate its molar conductivity. If Λ°m for acetic acid is 390.5 S cm² mol⁻¹, what is its dissociation constant?
Answer
Λm = κ × 1000/c = 7.896 × 10⁻⁵ × 1000/0.00241 = 32.76 S cm² mol⁻¹
α = Λm/Λ°m = 32.76/390.5 = 0.0839
Ka = cα²/(1 − α) = 0.00241 × (0.0839)²/(0.9161) = 1.85 × 10⁻⁵ mol L⁻¹
12. How much charge is required for the following reductions: (i) 1 mol of Al³⁺ to Al? (ii) 1 mol of Cu²⁺ to Cu? (iii) 1 mol of MnO₄⁻ to Mn²⁺?
Answer
(i) Al³⁺ + 3e⁻
⇒ Al: 3 F = 3 × 96487 = 2.89 × 10⁵ C
(ii) Cu²⁺ + 2e⁻
⇒ Cu: 2 F = 1.93 × 10⁵ C
(iii) MnO₄⁻ + 8H⁺ + 5e⁻
⇒ Mn²⁺ + 4H₂O (Mn: +7
⇒ +2): 5 F = 4.82 × 10⁵ C
13. How much electricity in terms of Faraday is required to produce (i) 20.0 g of Ca from molten CaCl₂? (ii) 40.0 g of Al from molten Al₂O₃?
Answer
(i) Ca²⁺ + 2e⁻
⇒ Ca; 40 g Ca needs 2 F
⇒ 20 g (0.5 mol) needs 1 F
(ii) Al³⁺ + 3e⁻
⇒ Al; 27 g Al needs 3 F
⇒ 40 g needs (3 × 40)/27 = 4.44 F
14. How much electricity is required in coulomb for the oxidation of (i) 1 mol of H₂O to O₂? (ii) 1 mol of FeO to Fe₂O₃?
Answer
(i) H₂O
⇒ ½O₂ + 2H⁺ + 2e⁻ (O: −2
⇒ 0): 2 F = 2 × 96487 = 1.93 × 10⁵ C
(ii) Fe²⁺
⇒ Fe³⁺ + e⁻ (one electron per FeO): 1 F = 96487 C ≈ 9.65 × 10⁴ C
15. A solution of Ni(NO₃)₂ is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?
Answer
Q = 5 × 20 × 60 = 6000 C. Ni²⁺ + 2e⁻
⇒ Ni (2 F deposit 58.7 g)
Mass of Ni = (58.7 × 6000)/(2 × 96487) = 1.825 g
16. Three electrolytic cells A, B, C containing solutions of ZnSO₄, AgNO₃ and CuSO₄, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?
Answer
Ag⁺ + e⁻
⇒ Ag: moles Ag = 1.45/108 = 0.01343 mol
⇒ Q = 0.01343 × 96487 = 1295.4 C
Time t = Q/I = 1295.4/1.5 = 863.6 s ≈ 14.4 min
Same charge flows through all cells (series). Cu²⁺ + 2e⁻
⇒ Cu: mass Cu = (63.5 × 1295.4)/(2 × 96487) = 0.426 g
Zn²⁺ + 2e⁻
⇒ Zn: mass Zn = (65.4 × 1295.4)/(2 × 96487) = 0.439 g
17. Using the standard electrode potentials given in Table 2.1, predict if the reaction between the following is feasible: (i) Fe³⁺(aq) and I⁻(aq) (ii) Ag⁺(aq) and Cu(s) (iii) Fe³⁺(aq) and Br⁻(aq) (iv) Ag(s) and Fe³⁺(aq) (v) Br₂(aq) and Fe²⁺(aq).
Answer
A reaction is feasible if E°cell = E°(cathode) − E°(anode) is positive.
(i) 2Fe³⁺ + 2I⁻
⇒ 2Fe²⁺ + I₂: E° = 0.77 − 0.54 = +0.23 V
⇒ feasible
(ii) 2Ag⁺ + Cu
⇒ 2Ag + Cu²⁺: E° = 0.80 − 0.34 = +0.46 V
⇒ feasible
(iii) 2Fe³⁺ + 2Br⁻
⇒ 2Fe²⁺ + Br₂: E° = 0.77 − 1.09 = −0.32 V
⇒ not feasible
(iv) Ag + Fe³⁺
⇒ Ag⁺ + Fe²⁺: E° = 0.77 − 0.80 = −0.03 V
⇒ not feasible
(v) Br₂ + 2Fe²⁺
⇒ 2Br⁻ + 2Fe³⁺: E° = 1.09 − 0.77 = +0.32 V
⇒ feasible
18. Predict the products of electrolysis in each of the following: (i) An aqueous solution of AgNO₃ with silver electrodes. (ii) An aqueous solution of AgNO₃ with platinum electrodes. (iii) A dilute solution of H₂SO₄ with platinum electrodes. (iv) An aqueous solution of CuCl₂ with platinum electrodes.
Answer
(i) Cathode: Ag⁺ + e⁻
⇒ Ag (silver deposited). Anode: Ag
⇒ Ag⁺ + e⁻ (silver anode dissolves). Net: silver transfers from anode to cathode; concentration of AgNO₃ stays the same.
(ii) Cathode: Ag⁺ + e⁻
⇒ Ag (Ag⁺ has higher reduction potential than water). Anode (inert Pt): 2H₂O
⇒ O₂ + 4H⁺ + 4e⁻ (NO₃⁻ is not oxidised). Products: Ag at cathode, O₂ at anode; solution becomes acidic (HNO₃).
(iii) Cathode: 2H⁺ + 2e⁻
⇒ H₂. Anode: 2H₂O
⇒ O₂ + 4H⁺ + 4e⁻ (in dilute acid water is oxidised in preference to SO₄²⁻). Products: H₂ at cathode, O₂ at anode.
(iv) Cathode: Cu²⁺ + 2e⁻
⇒ Cu (deposited). Anode: 2Cl⁻
⇒ Cl₂ + 2e⁻ (preferred over O₂ because of overpotential of oxygen). Products: Cu at cathode, Cl₂ at anode.
Important Extra Questions
What is the relation between E°cell, ΔrG° and the equilibrium constant K?
Answer
ΔrG° = −nFE°cell and ΔrG° = −RT ln K, so E°cell = (2.303RT/nF) log K = (0.059/n) log K at 298 K. A positive E°cell means negative ΔrG° and K > 1.
State Kohlrausch's law of independent migration of ions.
Answer
The limiting molar conductivity of an electrolyte is the sum of the limiting molar conductivities of its cation and anion, each multiplied by the number of such ions in the formula: Λ°m = ν₊λ°₊ + ν₋λ°₋.
Why is it not possible to measure the potential of a single electrode?
Answer
Oxidation and reduction always occur together, so a half-cell cannot work alone; only the potential difference between two electrodes can be measured. Hence potentials are measured against the standard hydrogen electrode, taken as 0.00 V.
Why does a mercury cell give a constant voltage throughout its life?
Answer
Its overall reaction, Zn(Hg) + HgO(s)
⇒ ZnO(s) + Hg(l), involves no ions whose concentration changes during its life, so the cell potential stays constant at about 1.35 V.
Give two methods of preventing corrosion of iron.
Answer
(i) Barrier protection by coating the surface with paint, oil or another metal (tin, chromium). (ii) Sacrificial protection, i.e. galvanisation with zinc or connecting to magnesium blocks, which are oxidised in preference to iron.