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 V at `25^(@)C` calculate the concentration of hydrogen ions at the positive electrode.

Monjur Alahi

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2025-01-04 04:42

1 Answers

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

Salim Ahmed Kawsar

The reaction is : `H^(+)+e^(-)rarr1//2H_(2)(g)`
`E_(cell)=(0.0591)/(1)"log"(C_(1))/(C_(2))`
`0.118 V=(0.0591)/(1)"log"(C_(1))/(C_(2))`
`"log"(C_(1))/(C_(2))=(0.118 Vxx1)/(0.0591)=2 or(C_(1))/(C_(2))="Antilog "2=10^(2)`
`C_(1)=C_(2)xx10^(2)=10^(-6)xx10^(2)=10^(-4) M`

5 views Answered 2023-02-05 15:29