Maharashtra BoardEnglish MediumSTD 12 ScienceChemistryElectrochemistry3 Marks
Question
Write electrode reactions for the electrolysis of aqueous $NaCl$.
✓
Answer
Reactions in electrolytic cell :
(i) Reduction half reaction at cathode : There are $Na ^{+}$and $H ^{+}$ions but since $H ^{+}$are more reducible than $Na ^{+}$, they undergo reduction liberating hydrogen and $Na ^{+}$are left in the solution.
$2 H _2 O _{(l)}+2 e ^{-} \rightarrow H _{2(g)}+2 OH _{(a q)}^{-}$(reduction) $E ^0=-0.83 V$ (ii) Oxidation half reaction at anode : At anode there are $Cl ^{-}$ and $OH ^{-}$. But $Cl ^{-}$ions are preferably oxidised due to less decomposition potential.
$2 Cl_{(a q)}^{-} \rightarrow 2 Cl_{(g)}+2 e^{-}$
$2 Cl_{(g)} \rightarrow Cl_{2(g)}$
$2 Cl_{(g)}^{-} \rightarrow Cl_{2(g)}+2 e^{-} \text {(overall reaction) }$
Net cell reaction: Since two electrons are gained at cathode and two electrons are released at anode for each redox step, the electrical neutrality is maintained. Hence we can write,
$2 H_2 O_{(i)}+2 e^{-} \rightarrow H_{2(g)}+2 OH_{(a q)} \text { (cathodic reduction) }$
$2 Cl _{\text {(aq) }}^{-} \rightarrow Cl _{2(g)}+2 e^{-}$(anodic oxidation)
$2 H_2 O_{(l)}+2 Cl_{(a q)}^{-} \rightarrow H_{2(g)}+2 OH_{(a q)}^{-}+Cl_{2(g)} \quad \text { (Overallcellreaction) }$
Since $Na ^{+}$and $OH ^{-}$are left in the solution, they form $NaOH _{( aq )}$.
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