Answer to Question #94169 in Physical Chemistry for Abdul sameer

Question

The emf of cell Cd |CdSO4(0.0093M) || CdSO4(xM)| Cd is 0.086V at 250C. Find the value of X.
Solution.

Answer 1

First half sell is |CdSO4(0.0093M)

potential is E₀ = -0.402 V

E = -0.402 + 0.0295log0.0093 = -0.4619 V

Full potential of cell is 0.086 V

Potential of second cell is the diffrence between full cell potential and first cell potential, than, if cell 2 is cathode

E₂= -0.3759 V

X = 7.669