The emf of the cell Zn|Zn2+(0.1 M)||Cd2+(M1)|Cd has been found to be 0.3305 V at 298 K. 

Calculate the value of M1, given Eo (zn2+/Zn)= -0.76 V, E o(Cd2+/Cd)= -0.40 V.

Dear student!
 

We have, E⁰_cell =  E⁰_(zn2+/Zn) -  E⁰_(Cd+2/Cd) = 0.76-0.40 = 0.36V

Further  E =0.3305, Here, C₂ = M₁ and  C₁ = 0.1M 

We know that, 

E= E⁰_cell - 0.0591/n log [C₂ ]/ [C₁] 

Here, no. of electrons involved (n)= 2

So, putting all the values we get , 0.3305 = 0.36 - 0.0591/2 log [M₁]/ [0.1]

or, 0.0295 = 0.0591/2 log [M₁]/ [0.1]

or,  0.0295 = 0.0295 log [M₁]/ [0.1]

or  log [M₁]- log[0.1] = 1

or,  log [M₁] = 1 + log[0.1] = 1-1 =0

So, [M₁] = antilog 0 = 1

Hence, concentration of M₁= 1M