Chemistry, asked by hulkyash1300, 9 months ago

Calculate the EMF of the given cell:
Al(s) l Al3+ (0.01M) ll Ni2+ (0.1M) ll Ni(s); if E0cell = 1.41V

Answers

Answered by rachnagupta2612
0

Answer:

Given Zn∣Zn

2+

(0.001M)∣∣Cu

2+

(0.1M)∣Cu

Overall cell reaction:

Zn⟶Zn

2+

+2e

Cu

2+

+2e

⟶Cu

Zn+Cu

2+

⟶Zn

2+

+Cu

E

cell

o

=standard reduction potential of cathode + standard oxidation potential of anode

E

cell

o

=0.34 to 0.76 V

E

cell

o

=1.1 V

K

C

=

[Cu

2+

]

[Zn

2+

]

=

10

−1

10

−3

=10

−2

EMF of the cell at any electrode concentration is:

E=E

o

n

0.059

log(K

C

)=1.1−

2

0.059

log(10

−2

)=1.1−

2

0.059

×(2)=1.1−0.059=1.041V

Answered by live4chess22
0

Answer:

Not possible..

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