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Calculate EMF of the following half cell...

Calculate `EMF` of the following half cells `:`
`a. Pt, H_(2)(2 atm)|HCl(0.02M)" "E^(c-)=0V`
`b. Pt , Cl_(2)(10 atm)|HCl(0.1 M)" "E^(-c)=1.36V`

Text Solution

Verified by Experts

`a. 2H^(o+)+2e^(-) rarr H_(2)(2 atm)`
`E_(cell)=(E^(c-)._(cell))-(0.059)/(2)log .(p_(H_(2)))/([H^(o+)])`
`=0-(0.059)/(2) log .(2)/((0.02)^(2))`
`=-(0.059)/(2) log .(2)/(4xx10^(-4))`
`=-(0.05)/(2)[log10^(4)-log2]`
`=-(0.059)/(2)[4-0.3]`
`=-(0.059)/(2)xx3.7=-0.0190V`
`E_(2H^(o+)|H_(2))=-0.019V`
`b .(` Take `0.059~~0.06)`
`Cl_(2)(10atm)+2e^(-)rarr 2Cl^(c-)(0.1M)`
`E_(Cl_(2)|2Cl^(c-))=E^(c-)._(Cl_(2)|2Cl^(c-))-(0.06)/(2) log .((0.1)^(2))/(10atm)`
`=1.36V-0.03log 10^(-3)`
`=1.36+0.03xx3=1.45V`
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