Question

Cd(S) / CdSO4 (XM) // CdSO4 (0.025M) / Cd(S). Emf of the given concentration cell at 

28 oC is 0.035V. Find the concentration of CdSO4 at anode. Given , R = 8.314J/K/mol, F = 96500​

Answer 1

Cd(S) / CdSO4 (XM) // CdSO4 (0.025M) / Cd(S). Emf of the given concentration cell at  

28 oC is 0.035V. Find the concentration of CdSO4 at anode. Given , R = 8.314J/K/mol, F = 96500​

Explanation:

Concentration cells are galvanic  cells made of two half-cells, each of which containing the same electrodes, but different concentration.

 EMF of a concentration cell is given by

E  cell  =  nF RT  ln  [C  1  ] [C  2  ]

​   =  0.0591 /n log  [C  1   ] /[C  2  ]

​  

 C₂ =0.025 M

C₁ =X

E  cell  =0.035 V

n=2

now,

0.035=  0.0591 /2 log 0.035 /X

X

0.035

​  

 or  X 0.035

​  

=antilog(1.1844)=15.29

or\\X=\frac{0.035}{15.29}=0.002289\ Mor

X=  

15.29

0.035

​  

=0.002289 M

So, concentration of\ \ CdSO_4  CdSO  

4

​  

 at anode = 0.002289\ M0.002289