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Q.
The mass of copper deposited from a solution of $CuSO_{4}$ by passage of $5 \,A$ current for $965$ second is (Mol. wt. of Copper = $63.5$)
Electrochemistry
Solution:
Current $(I) = 5 A$ and time $(t) = 965$ sec We know that equivalent weight of copper
=$\frac{Molecular\, weight}{Valancy}=\frac{63.5}{2}$
and quantity of electricity passed in coulomb
$=$ current $\times$ time $= 5 \times 965 = 4825 C.$ Since $96500$ coulombs
will deposit $\frac{63.5}{2}\,g$ of copper, therefore 4825 coulombs will deposit
$\frac{63.5 \times4825}{96500\times2}=1.5875 g$