1.05 g of lead ore containing impurity of Ag was dissolved in ${\rm{HN}}{{\rm{O}}_3}$ and the volume was made 350 mL. An Ag electrode was dipped in the solution. ${\rm{Pt}}\left( {{{\rm{H}}_2}} \right)\left| {{{\rm{H}}^ + }\left( {1{\rm{M}}} \right)} \right|\left| {{\rm{A}}{{\rm{g}}^ + }} \right|{\rm{Ag}}$. The${E_{{\rm{cell}}}}$ is 0.503 V at 298 K.The percent of lead in the ore is ($E_{{\rm{A}}{{\rm{g}}^ + }|{\rm{Ag}}}^ \circ = 0.80{\rm{V}}$)

NEET UG CHEMISTRY MCQ: ELECTRO CHEMISTRY

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1.05 g of lead ore containing impurity of Ag was dissolved in

\mathrm{HN}{\mathrm{O}_3}

and the volume was made 350 mL. An Ag electrode was dipped in the solution.

\mathrm{Pt}\left( {{\mathrm{H}_2}} \right)\left| {{\mathrm{H}^ + }\left( {1\mathrm{M}} \right)} \right|\left| {\mathrm{A}{\mathrm{g}^ + }} \right|\mathrm{Ag}

.
The

{E_{\mathrm{cell}}}

is 0.503 V at 298 K.The percent of lead in the ore is (

E_{\mathrm{A}{\mathrm{g}^ + }|\mathrm{Ag}}^ \circ = 0.80\mathrm{V}

)

View answer and explanation

Correct option: A

0.00033

Option A (0.00033) is correct because it is the calculated percentage of silver in the lead ore based on the given cell potential and standard electrode potential.

NEET UG CHEMISTRY MCQ: ELECTRO CHEMISTRY

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