Question

An acidic solution of $Cu^{2+}$ containing $0.4\, g$ of $Cu^{2+}$ ions is electrolysed until all the copper is deposited. What is the volume of oxygen evolved at $NTP$ ?

• A. $141\, cc$
• B. $31.75\, cc$
• C. $64\, cc$
• D. $32\, cc$

Answer 1

The Correct Option is A

\(H_{2}O \to2H^{+}+\frac{1}{2} O_{2}+2e^{-}\quad\) (Oxidation)
\(Cu^{2+}+2e^{-} \to Cu\quad\) (Reduction)
\(Cu^{2+}+H_{2}O\to Cu+2H^{+} +\frac{1}{2} O_{2}\) 
\(\frac{63.5}{2}=31.75\,g\) of \(Cu^{2+}\equiv\frac{1}{2}\times 22400=11200\,cc\) of \(O_{2}\)
\(31.75\,g\) of \(Cu^{2+}\) evolve \(11200 \,cc\) of \(O_{2}\) at \(NTP\)
\(0.4\,g\) of \(Cu^{2+}\) will evolve \(\frac{11200}{31.75}\times 0.4=141\,cc\) of \(O_{2}\)

So, the correct option is (A): 141 cc.

Answer 2

Moles of Cu \(= \frac {0.4}{63.5}= 6.3 \times 10^{-3}\) mol
Faraday used \(= 2 \times 6.3 \times 10^{-3} = 12.6 \times 10^{-3}\) mol
Now for oxygen,
\(2H_2O→O_2+4H^++4e^-\)
4 mole \(= 22400\ cm^3\) of \(O_2\).
\(12.6 \times 10^{-3} = \frac {22400}{2} \times 12.6 \times 10^{-3}\) cm³ of \(O_2\).
\(= 141\ cm^3\)

So, the correct option is (A): \(141\ cm^3\).