In the electrolysis of CuCl2 solution, the mass of cathode increased by 6.4g . What occurred at copper anode ?

Question

In the electrolysis of CuCl2 solution, the mass of cathode increased by 6.4g . What occurred at copper anode ? (A) 0.224litre of Cl2 was liberated (B) 1.12litre of oxygen was liberated (C) 0.05 mole Cu2 + passed into the solution (D) 0.1 mole Cu2 + passed into the solution

Tardigrade Admin · Accepted Answer

(CuCl)2(.aq.)leftharpoons(Cu)+ 2+2(Cl)- <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/c-xa3kk2llmexbmck2.jpg" /> At both electrode Cu will involve in the reaction Thus (geq)Anode =(geq)Cathode ⇒ n× 2=(6 . 4/64)× 2 ⇒ n=0.1mole of Cu+ 2 will form

Q. In the electrolysis of $C u C l_{2}$ solution, the mass of cathode increased by $6.4 g$ . What occurred at copper anode ?

$0.224 l i t re$ of $C l_{2}$ was liberated

$1.12 l i t re$ of oxygen was liberated

$0.05$ mole $C u^{2 +}$ passed into the solution

$0.1$ mole $C u^{2 +}$ passed into the solution

$(C u Cl)_{2} (a q) ⇌ (C u)^{+ 2} + 2 (Cl)^{-}$

At both electrode $C u$ will involve in the reaction
Thus $(g_{e q})_{A n o d e} = (g_{e q})_{C a t h o d e}$
$\Rightarrow n \times 2 = \frac{6.4}{64} \times 2$
$\Rightarrow n = 0.1 m o l e$ of $C u^{+ 2}$ will form

Q. In the electrolysis of CuCl2​ solution, the mass of cathode increased by 6.4g . What occurred at copper anode ?

0.224litre of Cl2​ was liberated

1.12litre of oxygen was liberated

0.05 mole Cu2+ passed into the solution

0.1 mole Cu2+ passed into the solution