A 1.050 g sample was dissolved in nitric acid to produce Pb2+(aq) and Ag+(aq). T
ID: 733103 • Letter: A
Question
A 1.050 g sample was dissolved in nitric acid to produce Pb2+(aq) and Ag+(aq). The solutionwas diluted to 355 mL with water, and Ag electrode was immersed in the solution, and the potential difference
between this electrode and S.H.E. was found to be 0.503 V. What is %Ag, by mass, in the Pb metal?
Explanation / Answer
For a general cell reaction: aA + bB ----> xX + yY in this case it is: 1/2H2 + Ag+ ----> H+ + Ag In the Ag/H2 cell we get the overall cell reaction: 1/2H2 | H+ || Ag+ | Ag where the hydrogen gas is being oxidised and the silver ions reduced. To calculate the concentrations in an electrochemical cell we need to apply the Nernst equation: E = Eºcell - RT/nF lnQ Where: R = gas constant; T = temp in Kelvin; F = Faraday constant; Q = reaction quotient; n = number of electrons transferred. The reaction quotient is: Q = [X]^x[Y]^y / [A]^a[B]^b As hydrogen gas, hydrogen ions and solid silver concentrations are at unity. We get Q = 1/[Ag+] Substituting into the Nernst equation, evaluating the constants and changing to Log base10 we get: 0.503 = 0.800 - 0.0592 log 1/ [Ag+] rearranging: log 1/[Ag+] = 5.017 1/[Ag+] = 104x10^3 [Ag+] = 9.615x10^-6 As the solution was only made up to 500cm^3 the final molar concentration is: 4.808x10^-6 moldm^3 converting this number moles into mass (Ar for silver = 107.87): 4.808x10^-6 x 107.87 = 5.186x10^-4g and finally calculate the percentage: 5.186x10^-4 x100 / 1.05 = 0.05%
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