Chemistry, asked by namrathameka8871, 1 year ago

An alloy of pb-ag weighing 1.08 g was dissolved in dilute hno3 and the total volume was made to be 100 ml. A silver electrode was dipped in the solution and the emf of the cell set up pt(s), h2(g) | h+(1m) || ag+ (aq) | ag(s) was 0.62 v. If eocell = 0.80 v, then what is the percentage of ag in the alloy?

Answers

Answered by knjroopa
6

Answer:

10

Explanation:

Given An alloy of pb-ag weighing 1.08 g was dissolved in dilute hno3 and the total volume was made to be 100 ml. A silver electrode was dipped in the solution and the emf of the cell set up pt(s), h2(g) | h+(1m) || ag+ (aq) | ag(s) was 0.62 v. If eocell = 0.80 v, then what is the percentage of ag in the alloy?

We know that

E cell = E o cell – 2.303 RT / nF log [ h+ / Ag]

E cell = Eo cell – 0.06 log [h+/ Ag]

0.62 = 0.80 – 0.06 log [ h+ / Ag]

-0.18 = -0.06 log 1 / Ag

 [ Ag+] = 10^-3 Molar

Weight of Ag = 10^-3 x 108

                        = 0.108 g

Weight % of Ag = 0.108 / 1.08 x 100

                           = 10%

Answered by ahnafv2005
0

Explanation:

e cell= e0cell - .06/1 log(h+/Ag+)

e cell =e0cell -.06log 1/(Ag+)

(Ag+)=10^-3M

volume= 100 ml which is .1L

so

weight of Ag=108×10^-3×10^-1 = .0108

percentage=.0108/1.08 ×100 =1

Similar questions