The EMF of the cell pr, H2 (1atm ) H+ / Kcl (1N) , Hg2cl2Hg, pt is 0.5164V at 25°c calculate the PH of solution if the reduction potential of normal calomel electrode is 0.28 V

Answer 1

To find the pH of the solution, we can use the Nernst equation:

Ecell = E°cell - (0.0592/n)log(Q)

where:
Ecell = cell potential = 0.5164 V
E°cell = standard cell potential = 0.28 V
n = number of electrons transferred in the overall reaction
Q = reaction quotient

First, let's write the overall cell reaction:

H2(g) + 2Hg2Cl2(s) -> 2Hg(l) + 4Cl-(aq)

Since there is no change in the number of moles of gas during the reaction, the reaction quotient Q simplifies to:

Q = [Cl-]^4

Next, we need to calculate [Cl-] from the given KCl concentration of 1N:

1N = 1 mol/L
Cl- comes from KCl in a 1:1 ratio, so [Cl-] = 1 mol/L

Now we can substitute the values into the Nernst equation:

0.5164 V = 0.28 V - (0.0592/2)log(1^4)
0.5164 V = 0.28 V - 0.0592log(1)
0.5164 V = 0.28 V

0.5164 V = 0.28 V

Solving for the term in the equation:

0.0592/2 * log(1) = 0

Therefore, at equilibrium the pH is 0.286 V.