Question

A solution of copper (II) sulphate (VI) is electrolysed between copper electrodes by a current of $10.0A$ for exactly $9650s$. Which remains unchanged?

• A. Cathodic plate
• B. Molar concentration of solution
• C. Anodic plate
• D. All of these

Answer 1

Answer: (B)

$CuS{O}_4\to C{u}^{2+}(aq)+S{O}_4^{2-}$

Electrons are made of copper. Thus Oxidatic

$Cu(s)⟶C{u}^{2+}(aq)+2e$

$C{u}^{2+}(aq)+2e^{-}⟶Cu(s)$

Reduction Copper formed $=w$

$w=zit=\frac{63.5\times 10.0\times 9650}{2\times 96500}g$

$=\frac{63.5}{2\times 63.5}=0.5mol$

$0.5$ mole of copper is dissolved from the anode. Thus, its mass decreases. $0.5$ mole of copper from the anode is deposited at the cathode. Thus, its mass increases. Thus, molar concentration of aqueous solution of $CuS{O}_4$ remains unchanged.