Question

Determination of ph of acid solution like 0.1 M HCl, 0.05 M HCl, 0.025 M HCl and 0.0125M HCl by using Quinhydrone and Calomel electrode.ComPare the value of ph with the theoretically calculated values.

Answer 1

Answer is-
The quinhydrone electrode potential at 25 degrees Celcius is 0.6992 V and for calomel electrode is 0.2438 V.
We can write the total potential as E = E(quin. or cal.) - E(2H+/H2).
In a respective solution, the potential of the hydrogen electrode will be
E = -0.059*lg(H+).
∴ E(quin.) = 0.6992 V - (-0.059*lg(H+))
or
 E(cal.) = 0.2438 V - (-0.059*lg(H+)).
To compare these calculations we need some practical results and we will get what we want.