Question

Calculate the electrode potentials of the following half cell at 25°C. 

Pt, Cl₂ (10 atm) | HCl (0.1 M)

Given: \[\ce{E^{\circ}_{\frac{1}{2}Cl_2/Cl^-}}\] = +1.36 V and \[\ce{E{^{\circ}_{H^+/\frac{1}{2}H_2}}}\] = 0.00 V.

Answer 1

Given: The half cell is Pt, Cl₂ (10 atm) | HCl (0.1 M)

Temperature = 25°C

`E_(1/2 Cl_2//Cl^-)^circ = +1.36` V

`E_(H^+//1/2 H_2)^circ = 0.00` V

The half reaction for chlorine reduction is \[\ce{Cl2_{(g)} + 2e- -> 2Cl^-_{ (aq)}}\]

The number of electrons transferred is n = 2

The nernst equation is E = `E^circ - 0.0591/2 log Q`    ...(i)

The reaction quotient `Q = ([Cl^-])/P_(Cl_2)`

Q = `(0.1)^2/10`

= `0.01/10`

Q = 0.001

Substitute this value of Q in the equation. (i)

E = `1.36 - 0.0591/2 log(0.001)`

log(0.001) = −3

E = `1.36 - 0.0591/2 (-3)`

= `1.36 + 0.1773/2`

= 1.36 + 0.08865

E = 1.44865 V