Question
Explain electrolysis of concentration aqueous solution of NaCl
✓
Answer
→ During the electrolysis of aqueous sodium chloride solution, the products are $NaOH , Cl _2$ and $H _2$.
→ In this case besides $Na ^{+}$and $Cl ^{-}$ions we also have $H ^{+}$and $OH ^{-}$ions along with the solvent molecules, $H _2 O$.
→ At the cathode there is competition between the following reduction reactions :
$
\begin{array}{l}
Na ^{+}( aq )+ e ^{-} \rightarrow Na ( s ) E _{\text {(cell) }}^{\ominus}=-2.71 V \\
H ^{+}( aq )+ e ^{-} \rightarrow \frac{1}{2} H _2(g) E _{\text {(cell) }}^{\ominus}=0.00 V
\end{array}
$
→ The reaction with higher value of $E ^{\ominus}$ is preferred and, therefore the reaction at the cathode during electrolysis is :
$
H ^{+}( aq )+ e ^{-} \rightarrow \frac{1}{2} H _2(g)...(1)
$
but $H ^{+}( aq )$ is produced by the dissociation of $H _2 O$ i.e.,
$
H _2 O (l) \rightarrow H ^{+}( aq )+ OH ^{-}( aq )
$...(2)
Therefore, the net reaction at the cathode may be written as the sum of (1) and (2) and we have
$
H _2 O ( l )+ e ^{-} \rightarrow \frac{1}{2} H _2(g)+ OH ^{-}
$
At the anode the following oxidation reactions are possible :
$
\begin{array}{l}
Cl ^{-}( aq ) \rightarrow \frac{1}{2} Cl _2(g)+ e ^{-} E _{\text {(cell) }}^{\ominus}=1.36 V \ldots \ldots . .(3) \\
2 H _2 O ( l ) \rightarrow O _2(g)+4 H ^{+}( aq )+4 e ^{-} E _{\text {(cell) }}^{\ominus}=1.23 V
\end{array}
$
→ The reaction at anode with lower value of $E ^{\ominus}$ is preferred and therefore, water should get oxidised in preference to $Cl ^{-}$(aq). However, on account of over potential of oxygen, reaction (3) is preferred. Thus, the net reactions may be summarised as:
$NaCl ( aq ) \xrightarrow{ H _2 O } Na ^{+}( aq )+ Cl ^{-}( aq )$
Cathode : $H _2 O (l)+ e ^{-} \rightarrow \frac{1}{2} H _2(g)+ OH ^{-}( aq )$
Anode : $Cl ^{-}( aq ) \rightarrow \frac{1}{2} Cl _2(g)+ e ^{-}$
Net reaction :
$
\begin{array}{l}
NaCl ( aq )+ H _2 O ( l ) \rightarrow Na ^{+}( aq )+ OH ^{-}( aq )+\frac{1}{2} H _2(g)+\frac{1}{2} Cl _2(g)
\end{array}$
→ In this case besides $Na ^{+}$and $Cl ^{-}$ions we also have $H ^{+}$and $OH ^{-}$ions along with the solvent molecules, $H _2 O$.
→ At the cathode there is competition between the following reduction reactions :
$
\begin{array}{l}
Na ^{+}( aq )+ e ^{-} \rightarrow Na ( s ) E _{\text {(cell) }}^{\ominus}=-2.71 V \\
H ^{+}( aq )+ e ^{-} \rightarrow \frac{1}{2} H _2(g) E _{\text {(cell) }}^{\ominus}=0.00 V
\end{array}
$
→ The reaction with higher value of $E ^{\ominus}$ is preferred and, therefore the reaction at the cathode during electrolysis is :
$
H ^{+}( aq )+ e ^{-} \rightarrow \frac{1}{2} H _2(g)...(1)
$
but $H ^{+}( aq )$ is produced by the dissociation of $H _2 O$ i.e.,
$
H _2 O (l) \rightarrow H ^{+}( aq )+ OH ^{-}( aq )
$...(2)
Therefore, the net reaction at the cathode may be written as the sum of (1) and (2) and we have
$
H _2 O ( l )+ e ^{-} \rightarrow \frac{1}{2} H _2(g)+ OH ^{-}
$
At the anode the following oxidation reactions are possible :
$
\begin{array}{l}
Cl ^{-}( aq ) \rightarrow \frac{1}{2} Cl _2(g)+ e ^{-} E _{\text {(cell) }}^{\ominus}=1.36 V \ldots \ldots . .(3) \\
2 H _2 O ( l ) \rightarrow O _2(g)+4 H ^{+}( aq )+4 e ^{-} E _{\text {(cell) }}^{\ominus}=1.23 V
\end{array}
$
→ The reaction at anode with lower value of $E ^{\ominus}$ is preferred and therefore, water should get oxidised in preference to $Cl ^{-}$(aq). However, on account of over potential of oxygen, reaction (3) is preferred. Thus, the net reactions may be summarised as:
$NaCl ( aq ) \xrightarrow{ H _2 O } Na ^{+}( aq )+ Cl ^{-}( aq )$
Cathode : $H _2 O (l)+ e ^{-} \rightarrow \frac{1}{2} H _2(g)+ OH ^{-}( aq )$
Anode : $Cl ^{-}( aq ) \rightarrow \frac{1}{2} Cl _2(g)+ e ^{-}$
Net reaction :
$
\begin{array}{l}
NaCl ( aq )+ H _2 O ( l ) \rightarrow Na ^{+}( aq )+ OH ^{-}( aq )+\frac{1}{2} H _2(g)+\frac{1}{2} Cl _2(g)
\end{array}$
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