Question

For the reaction

`2AgCl (s) + H_2 (g) ("1 atm") -> 2Ag (s) + 2H^(+)  (0.1 M) +  2Cl^(-) (0.1 M)`

`triangleG^0 = -43600 J at 25^@ C`

Calculate the e.m.f. of the cell

`[log 10^(-n) = -n]` 

Answer 1

`triangleG^0 = -nFE_(cell)^0`

`:. E_(cell)^0 = 0.226 V`

`E_(cell) = 0.226 - (0.059)/2 log (0.1)^2 (0.1)^2`

Therefore, `E_(cell) = 0.226 - (0.059)/2 log 10^(-4)`

= 0.226 + 0.059 x 2

= 0.344 V