Given that $2 C l^{-} \to C l_{2} + 2 e^{-}, E^{\circ} = - 1.36 V;$
$H_{2} O \to 2 H^{+} + \frac{1}{2} O_{2} + 2 e^{-}, E^{\circ} = - 1.23 V .$ On electrolysing 1M NaCl solution with inert electrodes, which of the following is correct regarding liberation of electrolytic products ?

Cathode : Na and $H_{2}$

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

Cathode : $N a, O_{2}$

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

Anode : $O_{2}$

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

Anode : $C l_{2}$

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

Open in App

Solution

The correct option is D
Anode : $C l_{2}$

Since $E_{\frac{N a^{+}}{N a}}^{\circ} = - 2.71 V$ < reduction potential of water (=- 0.83V), water will be reduced to $H_{2}$ and $O H^{-}$ at the cathode. Although
$E_{o x}^{\circ} (H_{2} O) = - 1.23 V > E_{\frac{C l^{-}}{C l_{2}}}^{\circ} = (- 1.36 V),$
Water will not be oxidised at the anode due to over potential of oxygen at platinum surface but $C l^{-}$ will oxidised to $C l_{2}$

Suggest Corrections

Similar questions

Given that $2 C l^{-} \to C l_{2} + 2 e^{-}, E^{\circ} = - 1.36 V;$
$H_{2} O \to 2 H^{+} + \frac{1}{2} O_{2} + 2 e^{-}, E^{\circ} = - 1.23 V .$ On electrolysing 1M NaCl solution with inert electrodes, which of the following is correct regarding liberation of electrolytic products ?

Q. Given are the reactions occuring during the electrolysis of sodium chloride solution:

$2 C l^{-} \to C l_{2} + 2 e^{-}$
$2 H_{2} O + 2 e^{-} \to H_{2} + 2 O H^{-}$

Which of the following is correct regarding liberation of electrolytic products?

Given are the reactions occuring during the electrolysis of sodium chloride solution:

$2 C l^{-} \to C l_{2} + 2 e^{-}$
$2 H_{2} O + 2 e^{-} \to H_{2} + 2 O H^{-}$

Which of the following is correct regarding liberation of electrolytic products?

Q. The half-cell reaction for the corrosion,

2 H^{+} + \frac{1}{2} O_{2} + 2 e^{-} \to H_{2} O; E^{\circ} = 1.23 V

F e^{2 +} + 2 e^{-} \to F e (s); E^{\circ} = - 0.44 V

Find the

△ G^{\circ}

(in kJ) for the overall reaction.

Q. Whats the potential of the reaction below given the half-reaction potentials:

2 F e_{2} + + C l_{2} \to 2 F e^{3 +} + 2 C l^{-}

F e^{3 +} + e^{-} \to F e^{2 +}; E = 0.77 V

C l_{2} + 2 e^{-} \to 2 C l^{-}; E = 1.36 V

Join BYJU'S Learning Program

Given that 2Cl−→Cl2+2e−,E∘=−1.36V; H2O→2H++12O2+2e−,E∘=−1.23V. On electrolysing 1M NaCl solution with inert electrodes, which of the following is correct regarding liberation of electrolytic products ?

Given that $2 C l^{-} \to C l_{2} + 2 e^{-}, E^{\circ} = - 1.36 V;$
$H_{2} O \to 2 H^{+} + \frac{1}{2} O_{2} + 2 e^{-}, E^{\circ} = - 1.23 V .$ On electrolysing 1M NaCl solution with inert electrodes, which of the following is correct regarding liberation of electrolytic products ?