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The observed emf of the cell, Pt|H(2) ...

The observed emf of the cell,
`Pt|H_(2) ("1 atm")|H^(+)(3xx10^(-4) M)||H^(+) (M_(1))|H_(2) ("1 atm")|Pt`
is 0.154 V. Calculate the value of `M_(1)` and pH of cathodic solution.

Text Solution

Verified by Experts

`E_(cell)=0.0591" log" M_(1)/(3xx10^(-4))`
or `"log"M_(1)/(3xx10^(-4))=0.154/0.0591=2.6058`
`M_(1)/(3xx10^(-4))=4.034xx10^(2)`
`M_(1)=4.034xx10^(2)xx3xx10^(-4)M`
`=0.121 M`
`pH=-log[H^(+)]=-log 0.121=0.917`
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