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Calculate the reduction potential of the...

Calculate the reduction potential of the following half - cells at `25^(@)C`, (1) `Pt|H_(2)("1.5 atm")|HCl"(0.01 M)"`
Given : `E_((1)/(2)Cl_(2)|Cl^(-))^(@)=+1.36V`

Text Solution

Verified by Experts

Reduction reaction at the given electrode is :
`2H^(+)(Aq)+2e rarr H_(2)(g)`
The Nernst equation for this reaction is
`E_("Red")=E_("Red")^(@)-(0.059)/(2)log.(p_(H_(2)))/([H^(+)]^(2))`
Here, `E_("Red")^(@)=0.0V` `therefore" "E_("Red")^(@)=0.0-(0.59)/(2)log.(.5)/((0.01)^(2))=-0.123V`
Hence, the reduction potential of the half - cell = `-0.123V`.
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