A) 25
B) 2.50
C) 10
D) 1
E) 50
Correct Answer: E
Solution :
\[Pt(s),{{H}_{2}}(g)|{{H}^{+}}(1M)||A{{g}^{+}}(aq)|Ag(s)\] EMF of cell \[=0.62\,V,E_{cell}^{o}=0.80\,V\] \[\begin{align} & \underline{\begin{align} & {{H}_{2}}\to 2{{H}^{+}}+2{{e}^{-}}(at\text{ }anode) \\ & 2A{{g}^{+}}+2{{e}^{-}}\to 2Ag(at\text{ }cathode) \\ \end{align}} \\ & {{H}_{2}}+2A{{g}^{+}}\to 2Ag+2{{H}^{+}}(overall\text{ }reaction) \\ \end{align}\] \[{{E}_{cell}}=E{}^\circ -\frac{2.303RT}{2F}\log \frac{{{[{{H}^{+}}]}^{2}}}{{{[A{{g}^{+}}]}^{2}}[{{H}_{2}}]}\] \[{{E}_{cell}}=E{}^\circ -\frac{2.303RT}{2F}\log \frac{1}{{{[A{{g}^{+}}]}^{2}}}\] \[0.62=0.80+\frac{2.303\times 0.06}{2}\log \frac{1}{{{[A{{g}^{+}}]}^{2}}}\] \[0.62=0.82+\frac{2\times 2.303\times 0.06}{2}\log [A{{g}^{+}}]\] \[-0.18=0.1382\log [A{{g}^{+}}]\] \[[A{{g}^{+}}]=0.05\,m\] \[\therefore \]Mole of\[A{{g}^{+}}\]in \[100\text{ }mL=0.05\times \frac{100}{1000}\times 108\] Wt. of\[A{{g}^{+}}\]in \[100\text{ }mL=0.05\times \frac{100}{1000}\times 108\] % of Ag in 1.08 g alloy \[=\frac{0.05\times 100\times 108}{1000\times 1.08}\times 100\] \[=50%\]
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