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
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%
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