An alloy of Pb-Ag weighing 1.08 g was dissolved in dilute H N O 3 and the volume made to 100 mL. A silver electrode was dipped in the solution and the emf of the cell set-up, Pt(s), H 2 ( g ) | H + ( 1 M ) | | A g + ( a q ) A g ( s ) was 0.62 V. If E c e l l o is 0.80 V, what is the percentage of Ag in the alloy? (At 25 ° C , RT/F = 0.06)

Step-by-step solution

Overall cell reaction is H 2 ( g ) + 2 A g + ⇌ 2 A g ( s ) + 2 H + ( a q ) E = E ° - 0.06 × 2.303 2 log ⁡ [ H + ] 2 [ A g + ] 2 p H 2 0.62 = 0.80 + 2 × 0.06 × 2.303 2 log ⁡ [ A g + ] [ A g + ] = 0.05 M Number of moles of A g + in 100 mL = M V 1000 = 0.05 × 100 1000 = 0.005 Mass of silver = 0.005 × 108 g Percentage of Ag in 1.08 g of alloy = 0.005 × 108 × 100 1.08 = 50 %