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Calculate the reduction potential of the following electrodes `:` `a.` `Pt,H_(2)(4 atm)|H_(2)SO_(4)(0.01M)` `b.` `Pt,H_(2)(1 atm)| HCl (0.2 M)` `c.` Calculate the potential of hydrogen electrode in contact with a solution whose `i pH=5" "ii. pOH=4` |
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Answer» `a.` Half cell reaction `:(n` factor for `H_(2)SO_(4)=2)` `H_(2)(4 atm)rarr 2H^(o+)(0.01M o r 0.02 N) +2e^(-)` `E_((H_(2)|2H^(o+)))=E^(c-)._((H_(2)|2H^(o+)))-(0.059)/(2)log.([H^(o+)])/(p_(H_(2)))` `0-(0.059)/(2)log.((0.02)^(2))/(4 atm)` `=-(0.059)/(2)log((0.02xx0.02)/(4))` `=-(0.059)/(2)log(10^(-4)) ` `=-(0.59)/(2)xx-4log 10` `=0.059xx2=0.118V` `E_((2H^(o+)|H_(2)))=-0.118V` `b.` Half cell reaction `:(n` factor for `HCl=1)` `H_(2)(1 atm) rarr2H^(o+)(0.2M o r 0.2 N)+2e` `E_((H_(2)|2H^(o+)))=E^(c-)._((H_(2)//2H^(o+)))-(0.059)/(2)log.([H^(o+)]^(2))/([p_(H_(2))])` `=0-(0.059)/(2)log.((0.2xx0.2))/(1 atm)` `=-(0.059)/(2)[ log 4xx10^(-2))]` `=-(0.059)/(2)[2 log 2-2]` `=(0.059)/(2)xx-1.4=0.059xx0.7=0.0413V` `E_((2H^(o+)|H_(2)))=-0.0413V` `c.` `i.` Use direct formula for the potential `(EMF)` of hydrogen electrode. `E_((EH^(o+)|H_(2)))=-0.059pH` `=-0.059xx5V` `=-0.295V` `ii.` `pOH=4" ":. pH=14-4=10` `E_((2H^(o+)|H_(2)))=-0.059pH` `=-0.059xx10V` `=-0.59V` |
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