Calculate the emf of the cell : `Pb(s)|Pb(NO_(3))_(2)(M_(1))||HCl(M_(2))|H_(2)(g)|Pt(s)`
When (i) `M_(1)=0.10 M`, `M_(2)=0.20 M` and `P_(H_(2))=1.00 "atm"`
(ii) `M_(1)=1.05" M"`, `M_(2)=1.0 M` and `P_(H_(2))=1.00" atm"`
(iii) `M_(1)=1.00 M`, `M_(2)=0.40" M"` and `P_(H_(2))=1.00" atm"`

Explanation

The overall cell reaction is : `Pb(s)+2H^(+)(M_(2))rarrPb^(2+)(M_(1))+H_(2)(g)`
According to the electrochemical series, `E_(Pb^(2+)//Pb)^(@)=-0.13 V " and " E_(H^(+)//H_(2))^(@)=0`
Let us calculate the emf of the cell in all the cases by appying Nernst equation
`E_(cell)=E_(cell)^(@)-(0.0591)/(2)"log"([Pb^(2+)(aq)])/([H^(+)(aq)]^(2))`
(i)` " " [Pb^(2)(aq)]=0.10 M, [H^(+)(aq)]=0.20 M`
`:. " "E_(cell)=0.00-(-0.13)-(0.0591)/(2)"log"(0.1)/((0.2)^(2))`
`E_(cell)=0.13-0.02955" log "(0.1)/(0.04)=0.13-0.02955 " log "2.5`
`=0.13-0.02955xx0.3979=0.13-0.01176=0.11824 V`
(ii)` " " [Pb^(2+)(aq)]=1.05 M,[H^(+)(aq)]=1.0 M`
`E_(cell)=0.00-(-0.13)-(0.0591)/(2)" log "(1.05)/((1.0)^(2))`
`=0.13-0.02955 " log " 1.05=0.13-0.02955xx0.0212`
=0.13-0.00063=0.12937 V
(iii)` " " [Pb^(2+)(aq)]=1.0M, [H^(+)(aq)]=0.40 M`
`E_(cell)=0.00-(-0.13)-(0.0591)/(2)" log "(1.0)/((0.40)^(2))`
`=0.13-0.02955xxlog6.25=0.13-0.02955xx0.7959`
=0.13-0.2352=0.10648 V