Question

The emf of Zn/Zn++

(0.01M) ||Zn++

(0.01M) |Zn cell is ––––––––––––– V

a) 1 c) 2

b) 0 d) None of the above​

Answer 1

Answer:

we have,

E=E0−20.0591​logFe2+Zn2+​

0.2905=E0−0.02955log0.0010.01​

Ecell0​=0.32

At equilibrium condition we have,

Ecell0​=20.0591​logK

0.02950.32​=logK

K=100.32/0.0295

Explanation: