Question

Number of electrons involved in the electrode position of $63.5$g of Cu from a solution of $CuS{O}_4$ is : $(N_A=6\times {10}^{23}$)

• A. $6\times {10}^{23}$
• B. $3\times {10}^{23}$
• C. $12\times {10}^{23}$
• D. $6\times {10}^{22}$

Answer 1

The correct option is C $12\times {10}^{23}$

In $CuS{O}_4$, oxidation number of $Cu$ is,

$x+(-2)=0$

$\Rightarrow x=+2$

$CuS{O}_4⟶Cu+S{O}_4^{2-}$

$\Rightarrow C{u}^{2+}+S{O}_4^{2-}⟶Cu+S{O}_4^{2-}$

$\Rightarrow C{u}^{2+}⟶Cu+2e^{-}$

i.e, $2e^{-}$ per mole of copper

Molar mass of $Cu=63.5g$

Then it would be $2e^{-}$ .