Question

If Cu²+/Cu electrode is diluted 100 times, then the change in emf is​

Answer 1

Answer:Here dear u have to apply 'Nernst Equation'.

I solved that ans.in Volt.

In MV it would be Decrease by 59 Mv

Option 2 is correct.

Best of luck for ur Boards dear❣

Answer 2

Let the initial concentration of the half cells be Co (for oxidised) and Cr (for reduced).

Say, after dilution, the concentration turns out to be Co' and Cr' respectively.

After dilution, the EMF of the cell is given by:

E = E° - 0.59(V/z)[log(Co'/Cr')

You see, Cu2+ is being diluted 100 times

Thus, Co'/Cr' = 100

=> log(Co'/Cr') = log[100]

=> log(Co'/Cr') = 2

and, Cu is being reduced to Cu2+ => z = 2

Hence,

E = E° - 0.59(1/2)*(log100)

E = E° = - 0.059

E - E° = -0.059 = -59 mV

Thus, the EMF of the cell decreases by 59mV.