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A cell contains two hydrogen electrode. ...

A cell contains two hydrogen electrode. The negative electrode is in contact with a solution of `10^(-6)` M hydrogen ions. The emf of the cell is 0.118 V at `25^(@)`. Calculate the concentration of hydrogen ions at the positive electrode.

Text Solution

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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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