Electrochemistry 1 Question 50
49. In an electrolysis experiment current was passed for $5 \mathrm{~h}$ through two cells connected in series. The first cell contains a solution of gold and the second contains copper sulphate solution. $9.85 \mathrm{~g}$ of gold was deposited in the first cell. If the oxidation number of gold is +3 , find the amount of copper deposited on the cathode of the second cell. Also calculate the magnitude of the current in ampere.
(Atomic weight of $\mathrm{Au}=197$ and atomic weight of $\mathrm{Cu}=63.5)$
(1983, 3M)
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Answer:
Correct Answer: 49. $(0.80 \mathrm{~A})$
Solution:
- Moles of $\mathrm{Au}$ deposited $=\frac{9.85}{197}=0.05$
$\Rightarrow$ gram equivalent of $\mathrm{Au}$ deposited $=0.05 \times 3=0.15$
Now, according to Faraday’s law of electrolysis, if same quantity of electricity is passed through different cells connected in series, same number of gram equivalents of electrolytes are discharged at respective electrodes.
$\Rightarrow$ gram equivalent of $\mathrm{Cu}$ deposited $=0.15$
$\Rightarrow$ amount of $\mathrm{Cu}$ deposited $=0.15 \times \frac{63.5}{2}=4.7625 \mathrm{~g}$
Also, Coulombs passed $=0.15 \times 96500=I \times 5 \times 60 \times 60$
$$ \Rightarrow \quad I=\frac{0.15 \times 96500}{5 \times 3600}=0.80 \mathrm{~A} $$
