In the electrolysis of aqueous sodium chloride solution which of the half-cell reaction will occur at the anode?
$N a_{(a q)}^{+} + e^{-} \to N a_{(s)}; E_{c e l l}^{o} = - 2.71$ V
$2 H_{2} O_{(l)} \to O_{2 (g)} + 4 H_{(a q)}^{+} + 4 e^{-}; E_{c e l l}^{o} = 1.23$ V
$H_{a q}^{+} + e^{-} \to \frac{1}{2} H_{2 (g)}; E_{c e l l}^{o} = 0.00$ V
$C l_{(a q)}^{-} \to \frac{1}{2} C l_{2 (g)} + e^{-}; E_{c e l l}^{o} = 1.36$ V

Question

In the electrolysis of aqueous sodium chloride solution which of the half-cell reaction will occur at the anode?
$N a_{(a q)}^{+} + e^{-} \to N a_{(s)}; E_{c e l l}^{o} = - 2.71$ V
$2 H_{2} O_{(l)} \to O_{2 (g)} + 4 H_{(a q)}^{+} + 4 e^{-}; E_{c e l l}^{o} = 1.23$ V
$H_{a q}^{+} + e^{-} \to \frac{1}{2} H_{2 (g)}; E_{c e l l}^{o} = 0.00$ V
$C l_{(a q)}^{-} \to \frac{1}{2} C l_{2 (g)} + e^{-}; E_{c e l l}^{o} = 1.36$ V

Toppr Admin · Accepted Answer

At anode- oxidation takes place-According to preferential discharge theory- more is the -x3B5-cell of anode potential more is the tendency for reaction to take place-hence- Cl2 gas is evolved-Cl-x229D-x27F6-12Cl2-e-x2212-x3B5-ocell-1-36 V

In the electrolysis of aqueous sodium chloride solution which of the half-cell reaction will occur at the anode?Na+(aq)+e−→Na(s);Eocell=−2.71V2H2O(l)→O2(g)+4H+(aq)+4e−;Eocell=1.23VH+aq+e−→12H2(g);Eocell=0.00VCl−(aq)→12Cl2(g)+e−;Eocell=1.36V

At anode, oxidation takes place.According to preferential discharge theory, more is the εcell of anode potential more is the tendency for reaction to take place.hence, Cl2 gas is evolved.Cl⊝⟶12Cl2+e−εocell=1.36 V

At anode, oxidation takes place.
According to preferential discharge theory, more is the $ε_{c e l l}$ of anode potential more is the tendency for reaction to take place.
hence, $C l_{2}$ gas is evolved.
$C l^{⊝} ⟶ \frac{1}{2} C l_{2} + e^{-} ε_{c e l l}^{o} = 1.36$ $V$