Question

After electrolysis of NaCl solution with inert electrodes for a certain period of time, 600 mL of the solution was left. Which was found to be 1 N in NaOH. During the same time, 31.75 g Cu deposited in the copper voltameter in series with the electrolytic cell. Calculate the percentage yield of NaOH obtained.

• A. $n=60$%
• B. $n=50$%
• C. $n=70$%
• D. None of these

Answer 1

The correct option is A $n=60$%

The number of equivalents of copper deposited is $\frac{31.8\times 2}{63.6}=1$
Hence, the number of equivalent of NaOH formed is 1.
The meq of NaOH formed is 1000.
But 600 mL of 1 N naOH is formed.
The experimetal yield or Meq of NaOH is $600\times 1=600$.
Hence, the percentage yield $=\frac{600}{1000}\times 100=60$ %