Question

The standard electrode potential of Pb ions is 0.126V. What will be the potential of a 0.5M solution at 298 K?

Answer 1

Answer:

We know that E

cell

is given by the equation,

E

cell

=E

cell

0

−

n

0.059

log(K)

Here for the given reaction,

Zn+Pb

2+

→Zn

2+

+Pb

n=2 for this reaction.

[Zn

2+

]=0.1M and [Pb

2+

]=1M

E

cell

0

=E

cathode

−E

anode

=0.126−(−0.763)=0.889V

Hence,

E

cell

=0.889−

2

0.059

log

1

0.1

=0.889V

Hence,option D is correct answer