Question

The no. of electrons lost or gained during electrolysis of 2 g $C{l}^{-}$ from NaCl(aq) to give $C{l}_2$ at anode is $x\times {10}^{20}$ electrons . $x$ is :

Answer 1

$2{Cl}^{-}\to 2e^{-}+{Cl}_2$

moles of ${Cl}^{-}=\frac{2}{35.5}=0.056moles$

moles of electron released for 0.056 moles of ${Cl}^{-}=0.05$

total number of electrons$=0.056\times 6.022\times {10}^{23}=339\times {10}^{20}$