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Following reactions may occur at carhode and anode during electrolysis of aqueous sodium chloride. What products will be held at anode and cathode? Use given E^{\circ} values to justify your answer.

Cathode :  Na^{+}(aq)+e^{-}\rightarrow Na (s), E^{\circ}= -2.71V

                  H^{+}(aq)+e^{-}\rightarrow \frac{1}{2}H_{2} (g), E^{\circ}=0.00V

Anode :    Cl^{-}(aq)\rightarrow \frac{1}{2}Cl_{2} (g)+e^{-}, E^{\circ}=+1.36V

                  2H_{2}O(aq)\rightarrow O_{2} (g)+4H^{+}+4e^{-}, E^{\circ}=+1.23V  

 

 

 

 
 
 
 
 

Answers (1)

Since reduction potential of hydrogen is greater, reduction of H^{+} tkes place at cathode.

Cathode :  H^{+}(aq)+e^{-}\rightarrow \frac{1}{2}H_{2}(g)

Since the oxidation potential of chlorine is greater than water, oxidation of Cl^{-} takes place at anode.

Anode : Cl^{-}(aq)\rightarrow \frac{1}{2}Cl_{2}(g)+e^{-}

Net cell reaction: H^{+}(aq)+Cl^{-}(aq)\rightarrow \frac{1}{2}H_{2}(g)+\frac{1}{2}Cl_{2}(g)

hence, H_{2} gas is formed at cathode and Cl_{2} gas is formed at anode.

Posted by

Sumit Saini

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