One litre of 0.5M KCl solution is electrolysed for 1 minute in a current of 16.08mA. Considering 100% efficiency, the pH of resulting solution will be a)7 b)9 c)8 d)10?

Question

Answer

Option c) is correct answer.

Answer

Electrolysis of KCl Solution

During the electrolysis of a KCl solution, the following reactions occur at the electrodes:

Anode (oxidation): 2Cl- → Cl2 + 2e-

Cathode (reduction): 2H2O + 2e- → H2 + 2OH-

Calculating the Amount of Charge

The current, I = 16.08 mA = 16.08 × 10^-3 A

The time, t = 1 minute = 60 seconds

Using the equation Q = I × t, we can calculate the amount of charge:

Q = (16.08 × 10^-3 A) × (60 s) = 964.8 C (Coulombs)

Calculating the Number of Moles of Cl2 Produced

From the balanced equation, we can see that 2 moles of electrons are required to produce 1 mole of Cl2:

2e- + 2Cl- → Cl2

Therefore, the amount of charge required to produce 1 mole of Cl2 is:

1 mole of Cl2 = 2 × Faraday's constant = 2 × 96,485 C/mol

The number of moles of Cl2 produced can be calculated using the equation:

n = Q/F

where n is the number of moles of Cl2, Q is the amount of charge, and F is Faraday's constant.

n = (964.8 C) / (2 × 96,485 C/mol) = 0.01 mol

Calculating the Concentration of OH-

Since Cl2 is produced at the anode, OH- ions will be formed at the cathode according to the balanced equation:

2H2O + 2e- → H2 + 2OH-

The number of moles of OH- ions formed will be equal to the number of moles of electrons used in the reduction reaction:

n(OH-) = 0.01 mol

The volume of the solution is given as 1 L. Therefore, the concentration of OH- ions can be calculated as:

[OH-] = n(OH-) / V(sol)

[OH-] = 0.01 mol / 1 L = 0.01 M

Calculating the pH of the Solution

Since the solution contains OH- ions, it will be basic in nature. The pH of a basic solution can be calculated using the equation:

pOH = -log[OH-]

pOH = -log(0.01) = 2

pH + pOH = 14 (at 25°C)

pH = 14 - pOH = 14 - 2 = 12

Conclusion

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