Question

5. Determine the emf of the galvanic cell given below
Ag/AgNO, (0.001 M) || AgNO, (0.1 M/Ag.
ЕО
Ag /Ag
0.80 V​

Answer 1

Answer:

Ag|AgI(0.05)M KI || (0.05)M AgNO

3

| Ag

K

sp

of AgI = [Ag

+

][I

−

]=[Ag

+

][0.05]

For the given cell

E

cell

=E

oxidised product(Ag)

+E

reduced product(Ag)

Using Nerst equation:

E

cell

=E

Ag/Ag

+

−

1

0.059

log[Ag

+

]

LHS

+E

Ag

+

/Ag

+

1

0.059

log[Ag

+

]

RHS

0.788=

1

0.059

log

[Ag

+

]

LHS

[Ag

+

]

RHS

0.788=0.059log

[Ag

+

]

LHS

0.05

[Ag

+

]

LHS

=2.203×10

−15

∴K

sp

(AgI)=2.203×10

−15

×0.05=1.10×10

−16