Question

When $96500$ coulomb of electricity is passed through a copper sulphate solution, the amount of copper deposited will be:

• A. $0.25mol$
• B. $0.50mol$
• C. $1.00mol$
• D. $2.00mol$

Answer 1

The correct option is A $0.25mol$

At the cathode

$C{u}_2\left(aq\right)+2e^{-}\to Cu\left(s\right)$

At the anode,

$4O{H}^{-}\left(aq\right)\to 2H_{2}O\left(l\right)+O_2\left(g\right)+4e^{-}$

Faraday’s constant $=96500\frac{C}{mol}$

To deposit 1 mole of copper, we need $2\times 96500C$

So, $96500C$ will deposit 0.5 moles of copper$=0.5\times 63.5=31.75g$

To liberate one mole of oxygen molecules,we need $4\times 96500C$

So, $96500C$ will liberate $=0.25$ moles