Question

In the electrolysis of aqueous sodium chloride solution which of the half cell reaction will occur at anode?

• A. $Na^+ (aq) +e^{-} \to Na(s)E^{\circ}_{\text{cell}}=2.71\,V$
• B. $2 H _{2} O (l) \longrightarrow O _{2}+4 H ^{+}+4 e^{-} E_{c e 11}^{\circ}=+1.23 V$
• C. $2 H ^{+}(a q)+ e ^{-} \longrightarrow \frac{1}{2} H _{2} E_{ cell }^{\circ}=0.00\, V$
• D. $Cl ^{-}(a q) \longrightarrow \frac{1}{2} Cl _{2}+e^{-} E_{ cell }^{\circ}=1.36 \,V$

Answer 1

Answer: (D)

In the electrolysis of aqueous sodium chloride solution, chlorine gas is liberated at the anode.
$
Cl _{( aq )}^{-}+ e ^{-} \rightarrow \frac{1}{2} Cl _{2}+ e ^{-} $
$ E _{\text {cell }}^{0}=1.36 V $
This is because $Cl ^{-}$ions have lower discharge potential than $OH ^{-}$ions.