Electrochemistry Numerical Problems for JEE and CBSE Exams

1. Electrolysis of Water:

• Volume of gases at STP: $$2H_{2}O(l)\to 2H_2(g)+O_2(g)$$ 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 $$Mass=Q/nF$$ $$Q=I\cdot t=10A\cdot 3600s=36,000C$$ For 1 mole of water: $$n=2(for{H}_2)+4(for{O}_2)=6$$ $$F=96500C/mol$$ $$Massof{H}_2=(36000C/6)\cdot (2g/mol)/(96500C/mol)$$ $$=1.2g{H}_2$$

  $$Massof{O}_2=(36000C/6)\cdot (32g/mol)/(96500C/mol)$$ $$=2.4g{O}_2$$

2. Electrode Potential:

• Zinc electrode potential: For the reaction $$Zn(s)\to Z{n}^{2+}(1M)+2e^{-}$$ The Nernst equation is: $$E=E°-\frac{RT}{nF}\ln [Z{n}^{2+}]$$ At standard conditions (25°C, 1 atm), $$E°=-0.76V$$, $$R=8.314J/molK$$, $$T=298K$$, $$n=2$$, $$F=96,500C/mol$$ And $$[Z{n}^{2+}]=1M$$

  Substituting the values, we get: $$E=-0.76V-\frac{8.314J/molK\cdot 298K}{2\cdot 96500C/mol}\cdot ln(1)$$ $$E=-0.76V$$ 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, $$C{u}^{2+}(1M)+2e^{-}\to Cu(s)$$ 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: $$Zn(s)|Z{n}^{2+}(1M)||C{u}^{2+}(1M)|Cu(s)$$ The cell potential is given by the Nernst equation: $$E_{cell} = E°{cell} - \frac{RT}{nF} \ln Q$$Atstandardconditions,$$E°{cell} = E°{Cu^{2+}/Cu} - E°{Zn^{2+}/Zn} = 0.34 V - (-0.76 V) = 1.10 V$$$$R = 8.314 J/mol K$$,$$T = 298 K$$,$$n = 2$$,,$$F = 96,500 C/mol$$

  Assuming the reaction proceeds in the forward direction, the reaction quotient Q is: $$Q=\frac{[C{u}^{2+}]}{[Z{n}^{2+}]}=\frac{1M}{1M}=1$$ Substituting the values into the Nernst equation, $$E_{cell}=1.10V-\frac{8.314J/molK\cdot 298K}{2\cdot 96500C/mol}\cdot \ln (1)$$ $$E_{cell}=1.10V$$ 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: $$W_{max}=-nF{E}_{cell}$$ Substituting the values, $$W_{max}=-2\cdot (96,500C/mol)\cdot (1.1V)=-212,300J=-212.3kJ$$ 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: $$2NaCl(aq)+2H_{2}O(l)\to 2NaOH(aq)+H_2(g)+C{l}_2(g)$$ 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: $$n=\frac{1gCu}{63.5g/mol}\times 2mol{e}^{-}/molCu$$ $$n=0.0314mol{e}^{-}$$ Substituting this value into Faraday’s law, we get: $$Q=0.0314mol{e}^{-}\times 96,500C/mol$$ $$Q\approx 3038.6C$$ 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: $$m=\frac{10A\times (1h\times 3600s/h)}{2mol{e}^{-}\times 96,500C/mol}$$ $$=\frac{36,000C}{193,000C/mol}$$ $$m\approx 0.186g$$ 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 lead-acid battery: The overall reaction for a lead-acid battery is: $$Pb(s) + PbO_2(s) + 2H_2SO_