CLASS 12 · CHEMISTRY · CHAPTER 2
Simple and explained answers for all 18 end-of-chapter questions on electrochemical cells, conductance, electrolysis, and batteries.
18 worked answers · short steps first, explanations when you need them
Textbook pages 57–60
Question 1
A metal with a more negative standard reduction potential is more easily oxidised and is the stronger reducing agent. The approximate activity order is Mg > Al > Zn > Fe > Cu. A metal displaces metals below it from their salt solutions.
Mg > Al > Zn > Fe > Cu (strongest to weakest displacement ability).
Question 2
Reducing power of a metal is greater when its standard reduction potential is more negative. Given potentials rise from K (−2.93 V), Mg (−2.37 V), Cr (−0.74 V), Hg (+0.79 V), to Ag (+0.80 V). Thus the reducing strength runs in the reverse order of these potentials.
Ag < Hg < Cr < Mg < K.
Question 3
Zinc is oxidised and silver ions are reduced: Zn(s) | Zn²⁺(aq) || Ag⁺(aq) | Ag(s). The zinc anode is negative. Electrons travel through the outer wire from zinc to silver; ions carry charge through the solutions and salt bridge. Anode: Zn → Zn²⁺ + 2e⁻. Cathode: 2Ag⁺ + 2e⁻ → 2Ag.
Cell: Zn(s) | Zn²⁺(aq) || Ag⁺(aq) | Ag(s). Negative electrode: Zn. External current is carried by electrons in the wire; ions move in the electrolyte/salt bridge. Half-reactions: Zn → Zn²⁺ + 2e⁻; 2Ag⁺ + 2e⁻ → 2Ag.
Question 4
(i) E° = E°Cd²⁺/Cd − E°Cr³⁺/Cr = (−0.40) − (−0.74) = +0.34 V; n=6. ΔG° = −nFE° = −6×96485×0.34 ≈ −1.97×10⁵ J mol⁻¹. log K = nE°/0.0591 ≈ 34.5, so K≈3.2×10³⁴. (ii) E° = 0.80−0.77 = +0.03 V; n=1. ΔG°≈−2.89 kJ mol⁻¹; log K=0.03/0.0591≈0.508, so K≈3.2.
Question 5
Use Ecell=E°cell−(0.0591/n)log Q. (i) 2.68 V; (ii) 0.529 V; (iii) 0.078 V; (iv) −1.29 V for the cell as written (reverse direction is spontaneous).
Use Ecell=E°cell−(0.0591/n)log Q.
Question 6
Zinc oxidation releases two electrons; silver oxide reduction uses two, so n=2. E°cell = E°(Ag₂O/Ag,OH⁻) − E°(Zn²⁺/Zn) = 0.344 − (−0.76) = 1.104 V. ΔrG°=−nFE°=−2×96485×1.104≈−2.13×10⁵ J mol⁻¹.
E°cell = 1.104 V; ΔrG° ≈ −213 kJ mol⁻¹.
Question 7
Conductivity κ measures how well a unit-sized portion of solution conducts electricity; its common unit is S cm⁻¹. Molar conductivity Λm is the conductivity contribution of the ions from one mole of electrolyte: Λm=κ×1000/c when c is in mol L⁻¹. On dilution, κ decreases because there are fewer ions per volume. Λm increases because ions move more freely; the increase is gradual for strong electrolytes and much larger for weak electrolytes.
Conductivity κ is conductance per unit dimensions of solution; molar conductivity Λm=κ×1000/c. Dilution lowers κ but raises Λm (slightly for strong and sharply for weak electrolytes).
Question 8
Use Λm=κ×1000/c for c in mol L⁻¹. Λm=0.0248×1000/0.20=124 S cm² mol⁻¹.
Λm = 124 S cm² mol⁻¹.
Question 9
Conductance G=1/R=1/1500 S. κ = cell constant × G, so cell constant=κ/G=κR. Cell constant=0.146×10⁻³×1500=0.219 cm⁻¹.
Cell constant = 0.219 cm⁻¹.
Question 10
Convert c from mol L⁻¹ to mol m⁻³ by multiplying by 1000. The table heading is 10²κ, so κ values are listed ×10⁻² S m⁻¹. Use Λm=κ/c. The resulting Λm values (S m² mol⁻¹) are: 0.01237, 0.01185, 0.011575, 0.011106, 0.010674. In S cm² mol⁻¹ these are 123.7, 118.5, 115.8, 111.1, and 106.7. Plot Λm against √c and extrapolate the nearly straight strong-electrolyte line to √c=0.
Λm values are 123.7, 118.5, 115.8, 111.1, and 106.7 S cm² mol⁻¹, respectively. Extrapolate the Λm vs √c plot to √c=0 to estimate Λm° (about 126 S cm² mol⁻¹).
Question 11
Λm=κ×1000/c=(7.896×10⁻⁵×1000)/0.00241=32.76 S cm² mol⁻¹. Degree of ionisation α=Λm/Λm°=32.76/390.5=0.0839. Ka=cα²/(1−α)=0.00241×(0.0839)²/(1−0.0839)=1.85×10⁻⁵ mol L⁻¹.
Λm≈32.8 S cm² mol⁻¹; Ka≈1.85×10⁻⁵ mol L⁻¹.
Question 12
Count the electrons in each balanced reduction; each mole of electrons carries 1 F≈96485 C. (i) Al³⁺+3e⁻→Al: 3F=2.89×10⁵ C. (ii) Cu²⁺+2e⁻→Cu: 2F=1.93×10⁵ C. (iii) MnO₄⁻+8H⁺+5e⁻→Mn²⁺+4H₂O: 5F=4.82×10⁵ C.
Question 13
For Ca²⁺+2e⁻→Ca, moles electrons = 2×(20.0/40.1)=0.998 mol e⁻. For Al³⁺+3e⁻→Al, moles electrons = 3×(40.0/27.0)=4.44 mol e⁻. One mole electrons is one Faraday.
Question 14
Oxidation of water: 2H₂O→O₂+4H⁺+4e⁻. Per mole water, 2 mol electrons are released: 2F. In FeO, Fe²⁺ becomes Fe³⁺, releasing one electron per Fe. One mole FeO therefore gives 1F. Convert F using 96485 C mol⁻¹.
Question 15
Cathode: Ni²⁺+2e⁻→Ni. Charge passed Q=It=5×(20×60)=6000 C. Ni mass = (M Q)/(nF) = (58.69×6000)/(2×96485)=1.82 g.
About 1.82 g Ni is deposited.
Question 16
Ag⁺+e⁻→Ag. Moles Ag=1.45/107.87=0.01344 mol, so Q=nF=1297 C. t=Q/I=1297/1.5=865 s ≈14.4 min. The same charge passes each cell. Cu²⁺+2e⁻→Cu gives mass=(63.55×Q)/(2F)=0.427 g. Zn²⁺+2e⁻→Zn gives mass=(65.38×Q)/(2F)=0.440 g.
Current flowed for about 865 s (14.4 min). Deposited masses: Cu≈0.427 g; Zn≈0.440 g.
Question 17
A proposed redox reaction is spontaneous when E°cell=E°(reduction at cathode)−E°(oxidation couple as reduction potential) is positive. (i) Fe³⁺ oxidises I⁻: E°≈0.77−0.54=+0.23 V, feasible. (ii) Ag⁺ oxidises Cu: E°≈0.80−0.34=+0.46 V, feasible. (iii) Fe³⁺ oxidises Br⁻: E°≈0.77−1.07=−0.30 V, not feasible. (iv) Ag cannot reduce Fe³⁺; E°≈0.77−0.80=−0.03 V, not feasible. (v) Br₂ oxidises Fe²⁺: E°≈1.07−0.77=+0.30 V, feasible.
Question 18
(i) Active silver anode dissolves to Ag⁺; Ag⁺ deposits as Ag at the cathode. (ii) Pt is inert: Ag⁺ deposits at cathode; water is oxidised at anode to O₂. (iii) Water is reduced to H₂ at cathode and oxidised to O₂ at anode; H₂SO₄ mainly provides conductivity. (iv) Cu²⁺ is reduced to Cu at cathode; chloride is oxidised to Cl₂ at the Pt anode.