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`

Explanation

`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`