Question

$20mL$ solution of $0.1M$ ferrous sulphate was completely oxidised using a suitable oxidising agent. What is the number of electrons exchanged?

• A. $1.204\times {10}^{22}$
• B. $193$
• C. $1930$
• D. $1.204\times {10}^{21}$

Answer 1

The correct option is D $1.204\times {10}^{21}$

During oxidation of $FeS{O}_4,{Fe}^{2+}$ ion is oxidises into ${Fe}^{3+}$ ion.

${Fe}^{2+}⟶{Fe}^{3+}+e^{-}$

i.e., $1e^{-}$ is transferred from $1$ molecule of $FeS{O}_4$,

i.e., from $1{Fe}^{2+}$ ion.

Now, molarity of $FeS{O}_4$ solution $=0.1M$ and volume of solution $=20mL=\frac{20}{1000}L$

$\therefore$ Number of moles $=0.1\times \frac{20}{1000}=0.002$

$\because$ In $1$ mole $FeS{O}_4$, number of molecules $=6.022\times {10}^{23}$

$\therefore$ In $0.002$ mole $FeS{O}_4$, number of molecules $=6.022\times {10}^{23}\times 0.002=1.204\times {10}^{21}$

$\therefore$ Number of electrons transferred $=1.204\times {10}^{21}$