A steady current of $$1.5\,A$$  flows through a copper voltameter for $$10\,min.$$  If the electrochemical equivalent of copper is $$30 \times {10^{ - 5}}g{C^{ - 1}},$$   the mass of copper deposited on the electrode will be

Solution :
If $$m$$ is the mass of a substance deposited or liberated on an electrode during electrolysis when a charge $$q$$ passes through electrolyte, then according to Faraday’s first law of electrolysis. Mass deposited is directly proportional to the charge flows.
i.e. $$m \propto q\,\,{\text{or}}\,\,m = Zq$$
where, $$Z$$ is a constant of proportionality and is called electrochemical equivalent (ECE) of the substance.
If an electric current $$i$$ flows through the electrolyte, then
$$m = Zit\,\,\left[ {q = it} \right]$$
Given, $$i = 1.5\,A,\,t = 10\,\min = 10 \times 60\,s,$$
$$Z = 30 \times {10^{ - 5}}g{C^{ - 1}}$$
Hence, mass of copper deposited on the electrode
$$\eqalign{ & m = 30 \times {10^{ - 5}} \times 1.5 \times 10 \times 60 \cr & = 27 \times {10^{ - 2}} = 0.27\,g \cr} $$