Calculate the minimum mass of NaOH required to be added in RHS to consume all the `H^(+)` present in RHS of the cell of emf + 0.701 volt at `25^(@)C` before its use. Also report the emf of the cell after addition of NaOH.
`Zn|underset("0.1 M")(Zn^(2+))||underset("1 litre")(HCl)|underset("1 atm")(Pt (H_(2)g)), E_(Zn//Zn^(2+))^(@)=0.760 V`

Explanation

The given cell is
`Zn| underset((0.1M))(Zn^(++)) || underset("1 lit")(HCl) | underset("1 atm")(H_(2)) |Pt`
So, the cell reaction can be written as
At anode : `Zn_((s)) to underset((0.01M))(Zn^(++) +2e^(-))`
At anode `2H^(+)+2e^(-) to underset(("1 atm"))(H_(2))`
`=Zn(S)+2H^(+) to underset((0.1M))(Zn^(+))+underset("1 atm")(H_(2)(g))`
`E_("cell")^(0)=E_(Zn//Zn^(+))^(0) +E_(H^(+)//H_(2))^(0)=(0.76+0)=0.76"volt"`
For Nearns.s equation
`E_("cell")=E_("cell")^(0)-(0.0592)/(2) log ([Zn^(++)] pH_(2))/([Zn(s)] [H^(+)]^(2))`
`0.701=E_("cell")^(0)-(0.0592)/(2) log ""([Zn^(++)])/([H^(+)]^(2)) =0.76 -(0.0592)/(2) log ""(0.1)/([H^(+)]^(2))`
`0.701 =E_("cell")^(0)-(0.0592)/(2) log ""([Zn^(++)])/([H^(+)]^(2))=0.76 -(0.0592)/(2) log ""(0.1)/([H^(+)]^(2))`
`0.701=0.76+(0.0592)/(2) log ""([H^(+)]^(2))/(0.1) therefore [H^(+)] =0.0316" mol lit"^(-1)`
or `[H^(+)] =0.0316` equivalent `lit^(-1)` [here mole=equivalent]
From law of equivalance
Equivalent of NaOH=Equivalent of HCl
`W/40=0.0316,` where W=Wt. of NaOH
W=1.264 gm
After addition of NaOH to cathode solution `[H^(+)]` becomes `10^(-7)`. Since acid and base are completely neutralised. Thus, new e.m.f. of cell.
`E_("cell")=E_("cell")^(0)+(0.0592)/(2) log ""([H^(+)]^(2))/(0.1)-0.760+(0.0592)/(2) log ((10^(-7))^(2))/(0.1)=0.3765V`