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  3. Chemistry
  4. On passing C ampere of current for time t sec through 1 L of 2 (M) CuSO4 solution (atomic weight of Cu = 63.5), the amount m of Cu (in gram) deposited on cathode will be

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Q. On passing $C$ ampere of current for time $t$ sec through $1\, L$ of $2\, (M)\,CuSO_{4}$ solution (atomic weight of $Cu \,= \,63.5$), the amount $m$ of $Cu$ (in gram) deposited on cathode will be

WBJEEWBJEE 2012Electrochemistry

Solution:

$m= \frac{E C t}{F}$
$=\frac{\frac{63.5}{2} \times C \times t}{96500}=\frac{31.75 \times C \times t}{96500}$