A cell constructed at 25C as follows. One hal-cell consist of the Cl2/CL- electr

Question

A cell constructed at 25C as follows. One hal-cell consist of the Cl2/CL- electrode with partial pressure of Cl2=.0050 atm and

[Cl-]= 2.00 M; the other involves the Cr2O7-/Cr3+ couple in acidic soultution in which [Cr2O7]=1.00 M, [Cr3+]=.800 M and pH=.5. Calculate the overall cell potential.  Equations given in picture. Thanks

A cell constructed at 25C as follows. One hal-cell consist of the Cl2/CL- electrode with partial pressure of Cl2=.0050 atm and [Cl-]= 2.00 M; the other involves the Cr2O7-/Cr3+ couple in acidic soultution in which [Cr2O7]=1.00 M, [Cr3+]=.800 M and pH=.5. Calculate the overall cell potential. Equations given in picture. Thanks

Explanation / Answer

Given equations are

1) Cl2 (g) + 2e- ---> 2Cl- Eo= 1.360

2) Cr207 + 14H+ + 6e- -----> 2Cr+3 + 7H20 E0=1.330

The balanced equation is given by

3 x 1 - 2

3 Cl2 + 2Cr+3 + 7H20 ----> 6 Cl- + Cr207 + 14 H+

Eo = Eo cathode - Eo anode

Eo = 1.360 -1.330

Eo= 0.03

By nernst equation

E= Eo - ( 0.0591 /n) log Q

here n =6 as 6 moles of electrons are transferred

E = E0 - ( 0.0591 /6) log [ ( Cl-) ^6 ( Cr207) ( H+) ^14 / (pCl2) ^3 (Cr+3) ^2 ]

Given pCl2 = 0.005

(Cr+3) = 0.8

( Cl-) = 2

( Cr207 ) = 1 M

Given pH = 5

pH =-log [H+]

-log [H+] = 5

[H+] = 10^-5

E = 0.03 - ( 0.0591/6 ) log [ 2^6 x ( 10^-5 )^14 / ( 0.005)^3 ( 0.8)^2 ]

E= 0.632

So Overall Cell potential is 0.632 V

When pH =0.5

-log [H+] = 0.5

[H+] = 10-0.5

E = 0.03 - ( 0.0591 /6) log [ 2^6 x ( 10^-0.5 )^14 / ( 0.005)^3 ( 0.8)^2 ]

E= 0.01125 V

So overall cell potential is 0.01125 V

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