During the electrolysis of copper (II) tetraoxosulphate (VI) solution, a steady current of 4.0 x 10\(^2\) A flowing for one hour liberated 0.48 g of copper....
Question 1 Report
During the electrolysis of copper (II) tetraoxosulphate (VI) solution, a steady current of 4.0 x 10\(^2\) A flowing for one hour liberated 0.48 g of copper. Calculate the mass of copper liberated by one coulomb of charge.
Answer Details
Data: current \(I = 4.0\times10^{2}\text{ A} = 400\text{ A}\), time \(t = 1\text{ hour} = 3600\text{ s}\), mass liberated \(m = 0.48\text{ g}\).
The total charge that passed is
\[ Q = It = 400 \times 3600 = 1.44\times10^{6}\ \text{C}. \]By Faraday's first law the mass liberated is proportional to the charge, so the mass liberated by one coulomb (the electrochemical equivalent, \(z\)) is
\[ z = \frac{m}{Q} = \frac{0.48}{1.44\times10^{6}} = 3.33\times10^{-7}\ \text{g C}^{-1}. \]Mass of copper liberated by one coulomb \(= 3.33\times10^{-7}\ \text{g}\) (i.e. \(3.33\times10^{-7}\) grams per coulomb).
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