Question

1.05 g of a lead ore containing impurity of Ag was dissolved in quantity of \[HN{{O}_{3}}\] and the volume was made 350 ml. A Ag electrode was dipped in the solution and potential of cell (\[{{E}_{cell}}\]) of
\[Pt({{H}_{2}})/{{H}^{+}}(1M)//A{{g}^{+}}/Ag\]
Was 0.0503 v at 298 K. Calculate % of Ag in the ore. \[{{E}_{A{{g}^{+}}/Ag}}=0.80v\].

Answer 1

Hint: From the data which is given in the question at anode oxidation (loss of electrons) takes place, at cathode reduction (gain of electrons) takes place.
The cell representation is\[Pt({{H}_{2}})/{{H}^{+}}(1M)//A{{g}^{+}}/Ag\]. From the cell representation we can say easily that anode is hydrogen electrode and cathode is silver electrode.
At anode: \[\dfrac{1}{2}{{H}_{2}}\to {{H}^{+}}(1M)+{{e}^{-}}\] (Oxidation)
At Cathode: \[A{{g}^{+}}(x)+{{e}^{-}}\to Ag\] (Reduction)
Total reaction = \[\dfrac{1}{2}{{H}_{2}}+A{{g}^{+}}(x)\to Ag+{{H}^{+}}(1M)\] (here n = 1)

Complete step by step answer:
 In the question tt is given that \[{{E}^{o}}\] = 0.80 V.
\[\begin{align}
  & Q=\dfrac{[{{H}^{+}}]}{[A{{g}^{+}}]}=\dfrac{1}{x} \\
 & E={{E}^{o}}-\dfrac{0.0591}{n}{{\log }_{10}}Q \\
 & 0.503=0.80-\dfrac{0.0591}{1}{{\log }_{10}}\left( \dfrac{1}{x} \right) \\
 & x=9.43\times {{10}^{-6}}M \\
 & \\
\end{align}\]
Q= it is the ratio of concentration of products to concentration of reactants.
n=number moles of the chemical involved in the reaction
x= concentration of silver ion
E=potential of the total cell
Number of moles of \[A{{g}^{+}}\]in 350 mL is
\[\begin{align}
  & =\dfrac{MV}{1000} \\
 & =\dfrac{9.43\times {{10}^{-6}}\times 350}{1000} \\
 & =3.3\times {{10}^{-6}} \\
\end{align}\]
M= concentration of silver ion
V= volume of the solution
Mass of Ag
\[\begin{align}
  & =3.3\times {{10}^{-6}}\times 108 \\
 & =3.56\times {{10}^{-4}}g \\
\end{align}\](108 is the atomic weight of the silver (Ag))
In the question it was mentioned that the solution contains 1.05 g of lead.
Therefore, the percentage of silver (Ag) in the ore =
\[\dfrac{3.56\times {{10}^{-4}}}{1.05}\times 100=0.0339%\]
So, the percentage of silver (Ag) in the ore is 0.0339%, nearly 0.031%.

Note: Don’t be confused with the words anode and cathode.
Anode is an electrode where oxidation (loss of electrons) of metal or chemical takes place.
Cathode is an electrode where reduction of metal (gain of electrons) or chemical takes place.