Question

calculate the emf of cell as asked in the photo

Answer 1

Here reaction occurs
Br– + H+ → 1/2Br2 + 1/2H2
e.g
Br– + e →1/2 Br2 { oxidation reaction }
Given,
E°(Br/Br–) = 1.08V
So, E°(Br–/Br) = -1.08V
Now, use Nernst equation
E = E° - 0.0591/nlog[Br–]^-1
= -1.08 - 0.0591/1 log( 10-²)-¹
= -1.08 - 0.0591× 2
= -1.1982 V

Again,
H+ + e → 1/2H2 { oxidation reaction }
E = E° - 0.0591/nlog[H+]-¹
E = 0 - 0.0591 ×log( 3 × 10-²)-¹
= -0.0591 × log( 100/3 )
= -0.0591 × ( 2 - log3)
= -0.0917 V

Now , EMF of cell = Ered + Eox
= (-1.1982 -0.0917 ) V
= -1.289 V