Question

The emf of the cell M /Mn+(0.02M)//H+(1M)/H2/Pt at 25°Cos 0.81V.calculate the valency of the metal if the SOP of the metal is 0.76V.

Answer 1

Answer: 2

Explanation:

$M+nH^+(1M)\rightarrow M^{n+}(0.02M)+nH_2$

Using Nernst equation :

$E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[M^{n+}]}{[H^{+}]^2}$

where,

The $E^0$ values have to be reduction potentials.

Thus for the metal which acts as anode ,the reduction potential is = -0.76 V

For hydrogen which acts as cathode ,the reduction potential is = 0 V

$E^0{cell}$ = standard electrode potential =$E^0{cathode}-E^0{anode}=0.00-(-0.76)=0.76V$

n = number of electrons in oxidation-reduction reaction

F = faradays constant = 96500 C

R= gas constant = 8.314 J/Kmol

T = temperature =$25^0c=25=273=298K$

Putting in the values:

$.81V=0.76-\frac{2.303\times 8.314\times 298}{n\times 96500}\log \frac{[0.02]}{[1]^2}$

$n=2$

Thus valency of metal is 2.