The correct option is B 0.06 V
The most important thing to note here is that at standard conditions. E∘=0 because we would have the same concentration of Zn2+ ions (unity) on both sides which is the only driving force in this case.
Also, flow of ions would be from concentrated solution to the dilute solution, which means that Zn2+ ions from the concentrated solution would be reduced and ones from dilute solution would get oxidised.
The half - cell reactions will be as follows:
Zn2+(2.4M)+2e−→Zn Reduction
Zn→Zn2+(0.024M)+2e− Oxidation
The Nernst equation for this reaction would be
Ecell=E∘−2.303RTnFlog([Zn2+(0.024M)][Zn(s)][Zn2+(2.4M)][Zn(s)])
Taking [Zn(s)] as unity, E_0 = 0 and plugging in n = 2, we end up with
Ecell=0.0592V
Rounding it off, we would get 0.06 V
The Zn2+ ions try to move from the concentrated half cell to a dilute solution. That driving force gives rise to 0.0592 V.