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.

A cell consists of two hydrogen electrodes. The negative electrode is

1.	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.
Answer» 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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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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