A) 99.97
B) 98.50
C) 0.033
D) 0.33
Correct Answer: A
Solution :
The cell reaction are: |
Anode: \[{{H}_{2}}\xrightarrow{{}}2{{H}^{+}}+2{{e}^{-}}\] |
Cathode: \[2A{{g}^{+}}+2{{e}^{-}}\xrightarrow{{}}2Ag\] |
Thus, \[{{E}_{cell}}=E_{O}^{{}^\circ }+E_{R}^{{}^\circ }+\frac{0.059}{2}{{\log }_{10}}\frac{{{[A{{g}^{+}}]}^{2}}{{P}_{{{H}_{2}}}}}{{{[{{H}^{+}}]}^{2}}}\] or \[0.503=0+0.80+\frac{0.059}{2}{{\log }_{10}}{{[A{{g}^{+}}]}^{2}}\] or \[[A{{g}^{+}}]=9.25\times {{10}^{-6}}M\] |
\[\therefore \]Mole of \[{{\operatorname{Ag}}^{+}}\] in 350mL\[=9.25\times {{10}^{-6}}\times \frac{350}{1000}\] |
\[\therefore \] Weight of\[{{\operatorname{Ag}}^{+}}\]in 350mL\[=9.25\times {{10}^{-6}}\times \frac{350}{1000}\times 108=3.497\times {{10}^{-4}}g\] |
\[\therefore \]\[%\]Of Ag in 1.05g alloy\[=\frac{3.497\times {{10}^{-4}}}{1.05}\times 100=0.033%\] |
\[\therefore \]\[%\]of lead in alloy \[=99.97%~\] |
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