During the electrolysis of the aqueous solution of copper sulphate using Pt electrodes, the reaction taking place at anode electrode is:

C u^{2 +} + 2 e^{-} \to C u

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C u \to C u^{2 +} + 2 e^{-}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 H_{2} O \to 4 H^{+} + O_{2} + 4 e^{-}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

H_{2} O + e^{-} \to O H^{-} + \frac{1}{2} H_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $2 H_{2} O \to 4 H^{+} + O_{2} + 4 e^{-}$
At anode oxidation of $H_{2} O$ occurs since the oxidation potential of $H_{2} O$ is greater than the oxidation potential of $S O_{4}^{2 -}$ ion.
Bubbles of gas (oxygen) are formed at the anode.
Cathode reaction: $C u^{2 +} (a q) + 2 e^{-} \to C u (s)$

Anode reaction: $2 H_{2} O (l) \to O_{2} (g) + 4 H^{+} (a q) + 4 e^{-}$

Suggest Corrections

Similar questions

O_{2} (g)

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} (c o r r e c t) C u (s) ⟶ & C u^{2 +} (a q) + 2 e^{-} E_{C u | C u^{2 +}}^{⊝} = - 0.34 V (w r o n g) 2 H_{2} O (l) ⟶ & O_{2} (g) + 4 H^{\oplus} (a q) + 4 e^{-} E_{H_{2} O | O_{2}}^{⊝} = - 1.23 V \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

\begin{matrix} I n (d) \end{matrix}

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} (c o r r e c t) N i (s) ⟶ & N i^{2 +} (a q) + 2 e^{-} E_{N i | N i^{2 +}}^{⊝} = - 0.25 V (w r o n g) 4 ⊝ O H (a q) ⟶ & O_{2} (g) + 2 H_{2} O (l) + 4 e^{-} E_{⊝ O H | O_{2}}^{\oplus} = - 0.4 V \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

C u S O_{4}

C u^{+ 2} + 2 e^{-} \to C u

C u \to C u^{+ 2} + 2 e^{-}

C u^{2 +} + 2 e^{-} ⟶ C u

E^{0} = 0.337 V

C u^{2 +} + e^{-} ⟶ C u^{+}

E^{0} = 0.153 V

E^{0}

C u^{+} + e^{-} ⟶ C u

C u S O_{4}

Join BYJU'S Learning Program