Question

The emf of the cell in which the following reaction Zn(s)+Ni+ (a =0.1)0 Zn2+ (a = 1.0) + Ni(s) occurs, is found to be 0.5105 V at 298 K. The standard e.m.f. of the cell is Chemistry

Answer 1

Answer: 0.54 V

Explanation:

$Zn+Ni^{2+}(0.1)\rightarrow Zn^{2+}(1.0)+2Ni$

Here Zinc undergoes oxidation by loss of electrons, thus act as anode. nickel undergoes reduction by gain of electrons and thus act as cathode.

Using Nernst equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}]}{[Ni^{2+}]}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E^o_{cell}$ = standard electrode potential = ?

$0.5105=E^o_{cell}-\frac{0.0592}{2}\log \frac{[1.0]}{[0.1]}$

$E_{cell}=0.54V$

Thus the standard e.m.f. of the cell is 0.54 V.