A current of 1 A is passed through 200ml solution of 0.05M .Licl solution for 96.5sec .Find out the pH of solution at the end of the electrolysis?

Question

Answer

Overview

In this question, we are required to find out the pH of a 200ml solution of 0.05M LiCl solution after passing a current of 1A for 96.5 seconds.

Calculations

First, we need to calculate the amount of charge that has passed through the solution:

Q = I * t

where I = 1A and t = 96.5s

Therefore, Q = 96.5C

Next, we need to calculate the number of moles of LiCl that have been electrolyzed:

n = Q / F

where F is the Faraday constant, which is 96,485 C/mol

Therefore, n = 0.001 mol

We also need to consider the reaction that takes place during electrolysis:

2H2O(l) → O2(g) + 4H+(aq) + 4e-

From the above reaction, we can see that for every 4 electrons that are transferred, 4 H+ ions are produced. Therefore, the concentration of H+ ions increases during electrolysis.

Now, we can calculate the new concentration of H+ ions:

mol H+ = 4 * n = 0.004 mol

volume of solution = 0.2 L

[H+] = mol H+ / volume of solution

Therefore, [H+] = 0.02 M

Finally, we can calculate the pH of the solution:

pH = -log[H+]

Therefore, pH = 1.7

Conclusion

The pH of the solution at the end of the electrolysis is 1.7. This is because the concentration of H+ ions increases during electrolysis, leading to a decrease in pH.

Answer

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