Shortcut Methods
Electrochemistry Numerical Problems for JEE and CBSE Exams
1. Electrolysis of Water:

Volume of gases at STP: $$ 2H_2O(l) → 2H_2(g) + O_2(g) $$ At STP, 1 mole of any gas occupies 22.4 L. So, when 1 mole of water is electrolyzed, 22.4 L of hydrogen and 11.2 L of oxygen are produced.

Mass of gases produced: Current (I) = 10 A, Time (t) = 1 hour = 3600 s $$Mass = Q/nF$$ $$Q = I \cdot t = 10 A \cdot 3600 s = 36,000 C$$ For 1 mole of water: $$n= 2 (for H_2) + 4 (for O_2) = 6$$ $$F= 96500 C/mol$$ $$Mass of H_2 = (36000 C /6) \cdot (2 g/mol)/ (96500 C/mol) $$ $$= 1.2 g H_2$$
$$Mass of O_2 = (36000 C/6) \cdot( 32 g/mol) / (96500 C/mol) $$ $$= 2.4 g O_2$$
2. Electrode Potential:

Zinc electrode potential: For the reaction $$Zn(s) → Zn^{2+}(1M) + 2e^$$ The Nernst equation is: $$E = E°  \frac{RT}{nF} \ln[Zn^{2+}]$$ At standard conditions (25°C, 1 atm), $$E°= 0.76 V$$, $$R= 8.314 J/mol K$$, $$T= 298 K$$, $$n=2$$, $$F= 96,500 C/mol$$ And $$[Zn^{2+}] = 1 M$$
Substituting the values, we get: $$E = 0.76 V  \frac{8.314 J/mol K \cdot 298 K}{2 \cdot 96500 C/mol} \cdot ln(1)$$ $$E = 0.76 V$$ Therefore, the electrode potential of a zinc electrode in a 1 M ZnSO4 solution is 0.76 V.

Standard electrode potential of copper: The given reaction is, $$Cu^{2+}(1M) + 2e^ → Cu(s)$$ By definition, the standard electrode potential is the potential of an electrode when the concentration of the reactants and products are at 1 M and the temperature is 25°C. Therefore, the standard electrode potential for the given reaction is 0.34 V.
3. Galvanic Cells:

Cell Potential: For the cell reaction: $$Zn(s)Zn^{2+}(1M)Cu^{2+}(1M)Cu(s)$$ The cell potential is given by the Nernst equation: $$E_{cell} = E°{cell}  \frac{RT}{nF} \ln Q$$ At standard conditions, $$E°{cell} = E°{Cu^{2+}/Cu}  E°{Zn^{2+}/Zn} = 0.34 V  (0.76 V) = 1.10 V$$ $$R = 8.314 J/mol K$$, $$T = 298 K$$, $$n = 2$$, , $$F = 96,500 C/mol$$
Assuming the reaction proceeds in the forward direction, the reaction quotient Q is: $$ Q= \frac{[Cu^{2+}]}{[Zn^{2+}]} = \frac{1M}{1M} = 1$$ Substituting the values into the Nernst equation, $$E_{cell} = 1.10 V  \frac{8.314 J/mol K \cdot 298 K}{2 \cdot 96500 C/mol} \cdot \ln (1)$$ $$E_{cell} = 1.10 V$$ Therefore, the cell potential of the given galvanic cell is 1.10 V.

Maximum electrical work: The maximum electrical work that can be obtained from the galvanic cell is given by: $$ W_{max} = nFE_{cell} $$ Substituting the values, $$W_{max} =  2 \cdot (96,500 C/mol) \cdot( 1.1 V) = 212,300 J = 212.3 kJ$$ Therefore, the maximum electrical work that can be obtained from the galvanic cell when 1 mole of zinc is oxidized is 212.3 kJ.
4. Electrolysis of Brine:

Mass of chlorine produced: Mass of NaCl in 100 g of brine solution = 20% of 100 g = 20 g Moles of NaCl = 20 g / 58.44 g/mol = 0.342 mol The balanced equation for the electrolysis of NaCl is: $$2NaCl(aq) + 2H_2O(l) → 2NaOH(aq) + H_2(g) + Cl_2(g)$$ Since 1 mole of NaCl produces 0.5 moles of chlorine gas, 0.342 mol of NaCl will produce 0.171 mol of chlorine gas. Mass of chlorine produced =0.171 mol * 71 g/mol = 12.1 g.

Volume of chlorine gas produced at STP: At STP, 1 mole of any gas occupies 22.4 L. Therefore, 0.171 mol of chlorine gas will occupy 0.171 * 22.4 = 3.8 L.
5. Faraday’s Law

Amount of charge required: The amount of charge required to electroplate 1 gram of copper can be calculated using Faraday’s law: $$Q = nF$$ Where: $$Q$$ is the charge in coulombs (C) $$n$$ is the number of moles of electrons transferred $$F$$ is Faraday’s constant (96,500 C/mol) For copper, 2 moles of electrons are required to electroplate 1 mole of copper. So, the number of moles of electrons transferred to electroplate 1 gram of copper is: $$ n = \frac{1 g Cu}{63.5 g/mol} \times 2 mol e^ / mol Cu$$ $$n = 0.0314 \ mol e^$$ Substituting this value into Faraday’s law, we get: $$ Q = 0.0314 mol e^ \times 96,500 C/mol$$ $$ Q ≈ 3038.6 C $$ Therefore, approximately 3039 C of charge is required to electroplate 1 gram of copper.

Mass of copper deposited: Using Faraday’s law, we can calculate the mass of copper deposited on the cathode: $$m = \frac{Q}{nF}$$ Where: $$m$$ is the mass in grams $$Q$$ is the charge in coulombs $$n$$ is the number of moles of electrons transferred $$F$$ is Faraday’s constant (96,500 C/mol) For copper, 2 moles of electrons are involved in the reaction. Substituting the given values, we get: $$m = \frac{10 A \times (1 h \times 3600 s/h)}{ 2 mol e^ \times 96,500 C/mol}$$ $$= \frac{36,000 C}{193,000 C/mol} $$ $$m ≈ 0.186 \ g$$ Therefore, approximately 0.186 g of copper will be deposited on the cathode when a current of 10 A is passed through a copper sulfate solution for 1 hour.
6. Batteries
 Emf of leadacid battery: The overall reaction for a leadacid battery is: $$Pb(s) + PbO_2(s) + 2H_2SO_