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

Options

A. `Na^(+)(aq)+e^(-)rarrNa(s),E_(cell)^(Theta)=-2.71V`
B. `2H_(2)O(l) rarrO_(2)(g)+4H^(+)(aq)+4e^(-)E_(cell)^(Theta)=123V`
C. `H^(+)(aq)+e^(-)rarr1/2H_(2)(g),E_(cell)^(-)=0.00V`
D. `Cl^(-)(aq)rarr1/2Cl_(2)(g)+e^(-),E_(cell)^(Theta)=1.36V`

Explanation

In case of electrolysis of aqueous NaCI oxidation reaction occurs at anode as follows
`Cl^(-)(aq)rarr1/2Cl_(2)(g)+e^(-) E^(o)=1.36V`
`2H_(2)O(l)rarrO_(2 ) (g)+4H^(+) (aq)+4e^(-)`
`E_(cell)^(@)=1.23V`
but due to lower `E_(cell)^(o)` value water should get oxidised in preference of `Cl^(-)(aq)` However, the actual reaction taking place in the concentrated solution of NaCl is (d) and not (b) i.e., `Cl_(2)` is produced and not `O_(2)` This unexpected result is explained on the basis of the concept of overvoltage i.e., water needs greater voltage for oxidation to `O_(2)` (as it is kinetically slow process) than that needed for oxidation of `Cl^(-)` ions to `Cl_(2)` thus, the correct option is (d) and not (b) 1. Which cell will measure standard electrode potential of copper electrode?
a) `Pt(s)|H_(2)(g,0.1"bar")|H^(+)(aq.,1M)"||"Cu^(2+)(a,1M)|Cu`
b) `Pt(s)|H_(2)(g,1"bar")|H^(+)(aq,1M)"||"Cu^(2+)(aq,2M)|Cu`
c) `Pt(s)|H_(2)(g,1"bar")|H^(+)(aq,1M)"||"Cu^(2+)(aq,1M)|Cu`
`Pt(s)| H_(2) (g,0.1"bar") | H^(+) (aq,0.1M)"||"Cu^(2+)(aq,1M)|Cu`
Ans (c) Standard electrode potential of copper electrode can be calculated by constructing concentration cell composed of two half cell reactions in which concentration of species on left hand and right hand side are unity. In such case cell potential is equal to standard electrode potential.
`underset ("Oxidationhalf cell reaction")ubrace(Pt(s)|H_(2)(g,1"bar")||H^(+)(aq, 1M)|)`
`underset("Reductionhalf cell reaction") ubrace(|Cu^(2+)(aq. 1M)|Cu)`