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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? |
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Answer» 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% |
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