The electrode reaction associated with the retention of blue colour of the electrolyte during electrolysis of $C u {S O}_{4}$ solution with copper electrodes 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^{-}

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

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

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

{S O}_{4}^{2 -} \to {S O}_{2} + O_{2} + 2 e^{-}

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

Open in App

Solution

The correct option is A $C u \to {C u}^{+ 2} + 2 e^{-}$
A copper deposit forms as the positive copper ions are attracted to the negative electrode (cathode) $C u^{2 +} (a q) + 2 e^{-} \to C u$ positive ion reduction by electron gain. The blue colour of the copper ion will fade as the copper ions are converted to the copper deposit on the cathode. But here The electrode reaction associated with the retention of blue colour of the electrolyte thus the reaction is $C u \to {C u}^{+ 2} + 2 e^{-}$ .

Suggest Corrections

Similar questions

C u S O_{4}

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

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

C u S O_{4}

C u_{(s)} \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

$C u - 2 e^{-} \to C u^{2 +}, E^{0} = - 0.347 V$

$S n - 2 e^{-} \to S n^{2 +}, E^{0} = + 0.143 V$

The EMF of the cell constructed with these electrodes is :

Join BYJU'S Learning Program