A cell contains two hydrogen electrodes. The negative electrode is in contact with a solution of `10^(-6) M` hydrogen ions. The emf of the cell is 0.1

Question

A cell contains two hydrogen electrodes. The negative electrode is in contact with a solution of `10^(-6) M` hydrogen ions. The emf of the cell is 0.1

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

Monjur Alahi

6 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

The cell may be represented as
`Pt|H_(2) ("1 atm")| underset(10^(-6) M)(H^(+))||underset("C M")(H^(+))|H_(2) ("1 atm")|Pt`
`{:(,),(,),(,):}`
`E_(cell)=0.0591/2"log"([H^(+)]_("Cathode")^(2))/([10^(-6)]^(2))`
`0.118=(0.0591)"log" ([H^(+)]_("Cathode"))/(10^(-6))`
`"log"([H^(+)]_("Cathode"))/10^(-6)=0.118/0.0591=2`
`([H^(+)]_("Cathode"))/10^(-6)=10^(2)`
`[H^(+)]_("Cathode")=10^(-6)xx10^(2)=10^(-4) M`