A) 0.94 V
B) 0.40V
C) 0.76 V
D) 0.20 V
Correct Answer: D
Solution :
\[{{H}_{2}}_{\left( g \right)}+2AgC{{l}_{\left( s \right)}}\to 2A{{g}_{\left( s \right)}}+2Cl{{-}_{\left( aq \right)}}+\,2\,{{H}^{+}}_{(aq)}\] \[{{E}_{cell}}=\,E{{{}^\circ }_{cell}}-\,\frac{0.06}{2}\log \,\,\left[ {{[{{H}^{+}}]}^{2}}\,{{[C{{l}^{-}}]}^{2}} \right)\] \[0.92=0+{{E}^{o}}_{AgCl/Ag,C{{l}^{-}}}-\frac{0.06}{2}\log \,\left[ {{({{10}^{-6}})}^{2}}\,{{({{10}^{-6}})}^{2}} \right]\]\[0.92+\frac{0.06}{2}\log \,\,{{10}^{-24}}\,=\,\,E{{{}^\circ }_{AgCl/Ag,C{{l}^{-}}}}\] \[E{{{}^\circ }_{AgCl/Ag,C{{l}^{-}}}}=\,\,0.20\]You need to login to perform this action.
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