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, E^{0}_{cell} = E^{0}_{(zn2+/Zn)} - E^{0}_{(Cd+2/Cd)} = 0.76-0.40 = 0.36V

Further E =0.3305, Here, C_{2} = M_{1} and C_{1} = 0.1M

We know that,

E= E^{0}_{cell} - 0.0591/n log [C_{2} ]/ [C_{1}]

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

So, putting all the values we get , 0.3305 = 0.36 - 0.0591/2 log [M_{1}]/ [0.1]

or, 0.0295 = 0.0591/2 log [M_{1}]/ [0.1]

or, 0.0295 = 0.0295 log [M_{1}]/ [0.1]

or log [M_{1}]- log[0.1] = 1

or, log [M_{1}] = 1 + log[0.1] = 1-1 =0

So, [M_{1}] = antilog 0 = 1

Hence, concentration of M_{1}= 1M

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