Question

The emf of cell Ag|AgSO4 (0.0083M) || AgSO4 (xM)| Ag is 0.074V at 250C. Find the value of X

Answer 1

Solution.

$E = \frac{0.059}{n} \times lg(\frac{C(cathode)}{C(anode)})$

$0.074 = \frac{0.059}{1} \times lg(\frac{X}{0.0083})$

X = 0.15 M

Answer:

X = 0.15 M