Chapter 5: Electrochemistry
Highly Detailed Mark-wise Solutions for Board Exam Preparation
1 Mark Questions (Very Short Answer)
Answer: Kohlrausch's law states that at infinite dilution, each ion migrates independently of its co-ion and makes its own definite contribution to the total molar conductivity of the electrolyte, irrespective of the nature of the other ion to which it is associated.
Answer: The SI unit of molar conductivity ($\Lambda_m$) is $\text{S m}^2 \text{ mol}^{-1}$ (Siemens meter squared per mole).
Commonly used unit: $\text{S cm}^2 \text{ mol}^{-1}$ or $\Omega^{-1} \text{cm}^2 \text{ mol}^{-1}$.
Answer: A salt bridge is a U-shaped glass tube containing a saturated solution of an inert strong electrolyte (such as $KCl$, $KNO_3$, or $NH_4NO_3$) in a gel-like substance (agar-agar), used to connect the two half-cells of a galvanic cell.
Answer: Cell constant ($b$) is the ratio of the distance between the electrodes ($l$) to the area of cross-section of the electrodes ($a$).
$$b = \frac{l}{a}$$
2 Mark Questions (Short Answer-I)
Answer:
Statement: The mass of any substance deposited or liberated at any electrode during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte.
Mathematical Expression:
$$W \propto Q$$
$$W = Z \cdot Q$$
Since $Q = I \times t$ (Current $\times$ time),
$$W = Z \cdot I \cdot t$$
Where $W$ = mass of substance, $Z$ = electrochemical equivalent, $I$ = current in Amperes, $t$ = time in seconds.
Answer:
- It provides an electrical contact between the two half-cells, thereby completing the inner electrical circuit.
- It maintains the electrical neutrality in both half-cell solutions by allowing the flow of opposite ions.
- It prevents the mixing of the two electrolytic solutions of the half-cells.
Answer:
Reference Electrode: An electrode whose potential is arbitrarily fixed or is exactly known at a given temperature, used to determine the potential of other unknown electrodes. (e.g., SHE).
Construction of SHE:
It consists of a platinum wire sealed in a glass tube. The lower end of the wire is attached to a small platinum foil coated with finely divided platinum black. This foil is immersed in a $1 \text{ M } H^+$ ion solution (e.g., $1 \text{ M } HCl$). Pure hydrogen gas at 1 atm pressure is continuously bubbled through the solution over the platinum foil at 298 K. Its standard potential is arbitrarily taken as exactly 0.00 V.
3 Mark Questions (Short Answer-II / Numericals)
Answer:
During discharging, the lead storage battery acts as a Galvanic cell, converting chemical energy into electrical energy.
1. At Anode (Oxidation):
Spongy lead is oxidized to $Pb^{2+}$ ions, which immediately combine with sulfate ions from the electrolyte ($H_2SO_4$) to form an insoluble precipitate of lead sulfate.
$$Pb(s) + SO_4^{2-}(aq) \rightarrow PbSO_4(s) + 2e^-$$
2. At Cathode (Reduction):
Lead dioxide ($PbO_2$) is reduced to $Pb^{2+}$ ions, which also combine with sulfate ions to form lead sulfate.
$$PbO_2(s) + 4H^+(aq) + SO_4^{2-}(aq) + 2e^- \rightarrow PbSO_4(s) + 2H_2O(l)$$
3. Net Cell Reaction:
$$Pb(s) + PbO_2(s) + 4H^+(aq) + 2SO_4^{2-}(aq) \rightarrow 2PbSO_4(s) + 2H_2O(l)$$
Or simply:
$$Pb(s) + PbO_2(s) + 2H_2SO_4(aq) \rightarrow 2PbSO_4(s) + 2H_2O(l)$$
Note: During discharging, sulfuric acid is consumed, and its density decreases.
$Mg(s) | Mg^{2+}(0.1 \text{ M}) || Cu^{2+}(0.001 \text{ M}) | Cu(s)$
Given: $E^\circ_{Mg^{2+}/Mg} = -2.37 \text{ V}$ and $E^\circ_{Cu^{2+}/Cu} = +0.34 \text{ V}$. March 2017, Oct 2020
Answer:
Step 1: Calculate Standard Cell Potential ($E^\circ_{cell}$)
$E^\circ_{cell} = E^\circ_{\text{Cathode}} - E^\circ_{\text{Anode}}$
Copper is cathode (right), Magnesium is anode (left).
$E^\circ_{cell} = 0.34 \text{ V} - (-2.37 \text{ V})$
$E^\circ_{cell} = 0.34 + 2.37 = 2.71 \text{ V}$
Step 2: Write Cell Reaction and find '$n$'
Anode: $Mg \rightarrow Mg^{2+} + 2e^-$
Cathode: $Cu^{2+} + 2e^- \rightarrow Cu$
Net: $Mg + Cu^{2+} \rightarrow Mg^{2+} + Cu$
Number of electrons transferred ($n$) = 2.
Step 3: Apply Nernst Equation
$$E_{cell} = E^\circ_{cell} - \frac{0.0592}{n} \log_{10} \frac{[Mg^{2+}]}{[Cu^{2+}]}$$
$$E_{cell} = 2.71 - \frac{0.0592}{2} \log_{10} \left( \frac{0.1}{0.001} \right)$$
$$E_{cell} = 2.71 - 0.0296 \log_{10} (100)$$
$$E_{cell} = 2.71 - 0.0296 \times 2$$
$$E_{cell} = 2.71 - 0.0592$$
$$E_{cell} = 2.6508 \text{ V}$$
The EMF of the cell is 2.6508 V.
4 Mark Questions (Long Answer / Combined)
Note: In the HSC Board Exam, 4-mark questions in Chemistry are frequently split into sub-questions (e.g., a 3-mark numerical combined with a 1-mark theory question).
(b) What are Galvanic cells? [1 Mark] March 2019, March 2021
Answer (a): Numerical on Faraday's Law
Given Data:
Current ($I$) = $1.5 \text{ A}$
Time ($t$) = $10 \text{ min} = 10 \times 60 \text{ seconds} = 600 \text{ s}$
Molar mass of Cu ($M$) = $63.5 \text{ g/mol}$
Faraday constant ($F$) = $96500 \text{ C/mol}$
Step 1: Calculate total charge ($Q$)
$Q = I \times t$
$Q = 1.5 \times 600 = 900 \text{ Coulombs}$
Step 2: Determine 'n' from reaction
Reduction of copper: $Cu^{2+} + 2e^- \rightarrow Cu(s)$
Here, 2 moles of electrons are required to deposit 1 mole of Cu. Thus, $n = 2$.
Step 3: Calculate Mass deposited ($W$)
Formula: $$W = \frac{M}{n \cdot F} \times Q$$
$$W = \frac{63.5}{2 \times 96500} \times 900$$
$$W = \frac{63.5 \times 900}{193000}$$
$$W = \frac{57150}{193000} \approx 0.296 \text{ g}$$
The mass of copper deposited is 0.296 g.
Answer (b): Theory
Galvanic (Voltaic) Cells: These are electrochemical devices that generate electricity through spontaneous chemical redox reactions. In these cells, chemical energy is converted into electrical energy. (e.g., Daniell Cell).
(b) The molar conductivity of 0.05 M $BaCl_2$ solution at 298 K is $223 \text{ S cm}^2 \text{ mol}^{-1}$. Calculate its conductivity. [2 Marks] Oct 2013, March 2022
Answer (a): Derivation
1. The maximum electrical work done by a galvanic cell is equal to the decrease in its Gibbs free energy.
$$-\Delta G = W_{\text{electrical}}$$
2. Electrical work done is the product of total charge transferred and the cell potential.
$$W_{\text{electrical}} = \text{Total Charge} \times \text{Cell Potential}$$
If $n$ moles of electrons are transferred, total charge $= nF$ (where $F$ is Faraday's constant).
$$W_{\text{electrical}} = nF \times E_{cell}$$
3. Therefore, $-\Delta G = nFE_{cell} \implies \Delta G = -nFE_{cell}$.
4. Under standard conditions (concentration 1 M, pressure 1 atm, temp 298 K):
$$\Delta G^\circ = -nF E^\circ_{cell}$$
Answer (b): Numerical on Conductivity
Given Data:
Molar conductivity ($\Lambda_m$) $= 223 \text{ S cm}^2 \text{ mol}^{-1}$
Concentration ($C$) $= 0.05 \text{ M} = 0.05 \text{ mol L}^{-1}$
Formula:
$$\Lambda_m = \frac{1000 \times \kappa}{C}$$
Where $\kappa$ is the conductivity.
Calculation:
$$\kappa = \frac{\Lambda_m \times C}{1000}$$
$$\kappa = \frac{223 \times 0.05}{1000}$$
$$\kappa = \frac{11.15}{1000} = 0.01115 \text{ S cm}^{-1}$$
The conductivity ($\kappa$) of the solution is $0.01115 \text{ S cm}^{-1}$.
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