Find EMF of cell `Pt |H_(2) | H^(+)(pH = 5) || H^(+) (pH =7) | H_(2)| Pt.`
Given that `(2.303 RT)/(F) = 0.06`

For the cell Pt_((s)), H_(2) (1 atm)|H^(+) (pH=2)||H^(+) (pH=3)|H_(2) (1atm),Pt The cell reaction is

For the cell Pt_((s)), H_(2) (1 atm)|H^(+) (pH=2)||H^(+) (pH=3)|H_(2) (1atm),Pt The cell reaction is

The emf of given cell Pt- H_2 | H^+|H_2 - Pt is :

Redox reactions play a pivotal role in chemistry and biology. The values of standard redox potential (E^(@)) of two half-cell reactions decide which way the reaction is expected to proceed. A simple example is a Daniel cell in which zinc goes into solution and copper gets deposited. For the cell: Pt |H_(2)(0.4"atm")H^(+)(pH=1)||H^(+)(pH=2)|H_(2)(0.1"atm")| Pt, the measured potential at 25^(@)C is: [Given: 2.303RT/F=0.06V, log 2=0.3]

For the cell Pt||H_2(0.4atm)|H^+(pH=1)||(H^+(pH=2)|H_2(0.1atm)|Pt The measured potential at 25^(@)C is

For the half-cell given Pt( H_(2) , 1 atm ) | H^+ pH = 2, the cell potential is :

For the half-cell given Pt( H_(2) , 1 atm ) | H^+ pH = 2, the cell potential is :

What will the EMF of the given cell be - Pt|H_(2)(P_(1))|H^(+)(aq)|H_(2)(P_(2))|Pt