Question

Calculate the weight of copper deposited by a quantity of electricity which displaces 5600 ml of O₂ at STP?

Answer 1

Dear User,
According to Faraday's Second Law of Electricity :
$$\begin{gathered} \frac{\mathrm{Weight}\mathrm{of}\mathrm{Cu}\mathrm{deposited}}{\mathrm{Weight}\mathrm{of}{\mathrm{O}}_2}=\frac{\mathrm{Equivalent}\mathrm{weight}\mathrm{of}\mathrm{Cu}}{\mathrm{Equivalent}\mathrm{weight}\mathrm{of}{\mathrm{O}}_2} \\ \mathrm{or},\frac{{\mathrm{w}}_{\mathrm{Cu}}}{{\mathrm{E}}_{\mathrm{Cu}}}=\frac{{\mathrm{w}}_{{\mathrm{O}}_2}}{{\mathrm{E}}_{{\mathrm{O}}_2}} \\ \mathrm{or},\frac{{\mathrm{w}}_{\mathrm{Cu}}}{31.77}=\frac{{\mathrm{w}}_{{\mathrm{O}}_2}}{{\displaystyle \frac{{\mathrm{M}}_{{\mathrm{O}}_2}}{4}}}(\mathrm{because}\mathrm{E}=\frac{\mathrm{Molecular}\mathrm{mass}}{\mathrm{Charge}\mathrm{on}\mathrm{its}\mathrm{ions}}\mathrm{and}2{\mathrm{H}}_2\mathrm{O}\to {\mathrm{O}}_2+4{\mathrm{H}}^{+}+4{\mathrm{e}}^{-}\mathrm{i}.\mathrm{e}.\mathrm{No}.\mathrm{of}\mathrm{electrons}=4) \\ \mathrm{or},\frac{{\mathrm{w}}_{\mathrm{Cu}}}{31.77}=\frac{{\mathrm{w}}_{{\mathrm{O}}_2}}{{\displaystyle {\mathrm{M}}_{{\mathrm{O}}_2}}}\times 4 \\ \mathrm{or},\frac{{\mathrm{w}}_{\mathrm{Cu}}}{31.77}={\mathrm{n}}_{{\mathrm{O}}_2}\times 4(\mathrm{since}\frac{{\mathrm{w}}_{{\mathrm{O}}_2}}{{\displaystyle {\mathrm{M}}_{{\mathrm{O}}_2}}}={\mathrm{n}}_{{\mathrm{O}}_2}) \\ \mathrm{Now},\mathrm{no}.\mathrm{of}\mathrm{moles}\mathrm{of}{\mathrm{O}}_2,{\mathrm{n}}_{{\mathrm{O}}_2}=\frac{5600}{22400}=\frac{1}{4} \\ \mathrm{Therefore},{\mathrm{w}}_{\mathrm{Cu}}=\frac{1}{4}\times 4\times 31.77=31.77\mathrm{g} \end{gathered}$$