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InterviewSolution
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A cell consists of two hydrogen electrodes. The negative electrode is in contact with a solution having 10^(-6)MH^(+) ion concentration. Calculate the concentration of H^(+) ions at the positive electrode, if the emf of the cell is found to be 0.118 V at 298 K. |
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Answer» Solution :Here, `C_(1)=10^(-6)M,""C_(2)=`tobe CALCULATED. For given concentration CELL, `E_(cell)=(0.0591)/(n)"log"(C_(2))/(C_(1))` `0.118=(0.0591)/(1)"log"(C_(2))/(10^(-6))" or ""log"(C_(2))/(10^(-6))=2" or "(C_(2))/(10^(-6))="Antilog 2"=10^(2) ""thereforeC_(2)=10^(2)xx10^(-6)=10^(-4)M`. |
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