Michael Faraday was the first scientist who described the quantitative aspects of electrolysis. Now Faraday’s laws also flow from what has been discussed earlier. Faraday’s Laws of Electrolysis After his extensive investigations on electrolysis of solutions and melts of electrolytes, Faraday published his results during 1833-34 in the form of the following well known Faraday’s two laws of electrolysis:
(i) First Law: The amount of chemical reaction which occurs at any electrode during electrolysis by a current is proportional to the quantity of electricity passed through the electrolyte (solution or melt).
(ii) Second Law: The amounts of different substances liberated by the same quantity of electricity passing through the electrolytic solution are proportional to their chemical equivalent weights (Atomic Mass of Metal ÷ Number of electrons required to reduce the cation).
The density of Cu is 8.94 g/cm3. What is the quantity of electricity needed to plate an area 10 cm × 10 cm to a thickness of 10−2 cm using CuSO4 solution?
(Cu = 63.5 g mol−1, 1 F = 96500 C mol−1)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 27172 C
By Faraday’s first law, the mass of copper deposited is proportional to the quantity of electricity passed. In \( \ce{CuSO4} \) solution each \( \ce{Cu^2+} \) ion takes two electrons at the cathode:
\( \displaystyle \ce{Cu^2+(aq) + 2e^- -> Cu(s)} \)
So one mole of copper needs 2 F of charge. First find the mass of the copper layer from its volume and density:
\( \displaystyle V=10\,\mathrm{cm}\times10\,\mathrm{cm}\times10^{-2}\,\mathrm{cm}=1\,\mathrm{cm^3} \)
\( \displaystyle m=V\times d=1\,\mathrm{cm^3}\times8.94\,\mathrm{g\,cm^{-3}}=8.94\,\mathrm{g} \)
Moles of copper:
\( \displaystyle n(\ce{Cu})=\frac{8.94\,\mathrm{g}}{63.5\,\mathrm{g\,mol^{-1}}}=0.1408\,\mathrm{mol} \)
Quantity of electricity:
\( \displaystyle Q=n(\ce{Cu})\times2\times F \)
\( \displaystyle Q=\frac{8.94}{63.5}\times2\times96500\,\mathrm{C} \)
\( \displaystyle Q=27172\,\mathrm{C} \)