Chemistry, asked by ushauhegde1973, 3 months ago

1C An electrochemical cell consists of Cu electrode dipped in 0.5M CuSO4 solution and Ag
electrode dipped in 0.25M AgNO3 solution. Write the cell representation, cell reactions
& calculate the EMF. Given SEP's of Cu & Ag are 0.34V & 0.8V respectively. ​

Answers

Answered by 6715devarshdaryani
0

Answer:

Given, E

Cu

2+

/Cu

=0.337 volt and E

Ag

+

/Ag

=0.799 volt. The standard emf will be positive if Cu/Cu

2+

is anode and Ag

+

/Ag is cathode. The cell can be represented as:

Cu∣Cu

2+

∥Ag

+

∣Ag

The cell reaction is,

Cu+2Ag

+

→Cu

2+

+2Ag

E

cell

= Oxid. potential of anode + Red. potential of cathode

=−0.337+0.799

=0.462 volt

Applying the Nernst equation,

E

cell

=E

cell

2

0.0591

log

[Ag

+

]

2

[Cu

2+

]

When, E

cell

=0

E

cell

=

2

0,0591

log

[Ag

+

]

2

[Cu

2+

]

or log

[Ag

+

]

2

[Cu

2+

]

=

0.0591

0.462×2

=15.6345

[Ag

+

]

2

[Cu

2+

]

=4.3102×10

15

[Ag

+

]

2

=

4.3102×10

15

0.01

=0.2320×10

−17

=2.320×10

−18

[Ag

+

]=1.523×10

−9

M.

Similar questions