Question

(a) calculate EMF of following cell:
Zn(s)/Zn^2+(0.1 M) || (0.01M) Ag+ / Ag (s)
Given : E° zn2+ / zn = -0.76 V,
E° Ag+/Ag = 0.80 V,
log 10 =1
(b) X and Y are two electrolytes, on dilution molar conductivity of'X' increases 2.5 times while that 'Y' increases 25 times .which of the two is weak electrolyte and why?

Answer 1

Answer:

(a) E = 1.37

Explanation:

E cell = 0.80 -(-0.76)

=1 .46

than, E = 1.46 - 0.059/2 log 0.1/(0.01)^2

= 1.46 - 0.0295 log 1000

= 1.46-3(0.0295)

= 1.371

Answer 2

a) EMF of the cell = 1.471 V

b) Y is a weak electrolyte, because it has steep increase in its molar conductivity on dilution.

• Given, we have  two half cells,

• Ag+1(aq)  + e-1  --->  Ag(s)  reduction  Eo = 0.80 V
• Zn+2(aq)  + 2e-1  ---> Zn(s)  oxidation  Eo = -0.76
• The net cell reaction is  2 Ag+1(aq) + Zn(s) ---- > 2Ag(s)  + Zn+2(aq)
• As, reduction potential of Zn is lower, it undergoes oxidation.
• We know, Ecell = Eo reduction - Eo oxidation  = 0.80 - (-0.76)  = 1.56 V
• Now. we know    EMF = Eo cell - (0.0591 / n) log[anode]/[cathode], where n is number of electrons transferred
• EMF = 1.56 - (0.0591 /2) log0.1/(0.01)^2 = 1.56 - (0.0591 /2) log10^3
• EMF = 1.56 - (0.0591/2) *3  = 1.56 - 0.08865  = 1.471 V

• b) Strong electrolyte has already completely dissociated in the solution, so on dilution only less amount of its molar conductivity increases.
• Whereas ,for weak dilution on dilution its dissociation increases, leading to steep increase in its molar conductivity.