COMEDK · Chemistry · 18. Electrochemistry
When a quantity of electricity is passed through \(\mathrm{CuSO}_{4}\) solution, \(0.16 \mathrm{~g}\) of copper gets deposite(d)If the same quantity of electricity is passed through acidulated water, then the volume of \(\mathrm{H}_{2}\) liberated at STP will be [Given, atomic weight of \(\mathrm{Cu}=64\) ]
- A \(4.0 \mathrm{~cm}^{3}\)
- B \(56 \mathrm{~cm}^{3}\)
- C \(604 \mathrm{~cm}^{3}\)
- D \(8.0 \mathrm{~cm}^{3}\)
Answer & Solution
Correct Answer
(B) \(56 \mathrm{~cm}^{3}\)
Step-by-step Solution
Detailed explanation
According to second law of Faraday, \[ \begin{aligned} \frac{w_{1}}{E_{1}} &=\frac{w_{2}}{E_{2}} \\ \frac{0.16}{64} &=\frac{w_{2}}{2} \\ w_{2} &=\frac{0.16 \times 2}{64}=\frac{0.01}{2}=0.005 \mathrm{~g} \end{aligned} \] Volume of…
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