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 cell reaction is,
`Zn+2H^(+) rarr Zn^(2+) +H_(2)`
Applying Nernst equation,
`E_(cell)=E^(@)-0.0591/2"log"([Zn^(2+)])/([H^(+)]^(2))`
`0.701=0.760-0.0591/2"log" ([Zn^(2+)])/([H^(+)]^(2))`
So, `"log"([Zn^(2+)])/([H^(+)]^(2))=(0.0591xx2)/0.0591=2`
or `([Zn^(2+)])/([H^(+)]^(2))=10^(2)`
`[H^(+)]^(2)=0.1/10^(2)=10^(-3)`
`[H^(+)]=0.0316" mol L"^(-1)`
Thus, 0.0316 mol/litre of NaOH is required to neutralise `H^(+)` ions.
Mass of NaOH = `0.0316xx` Mol. mass of NaOH
`=0.0316xx400=1.264 g`
After addition of NaOH, the solution becomes neutral, i.e., the concentration of `H^(+)` ions in cathodic solution becomes `10^(-7)`.
Applying again Nernst equation,
`E_(cell)=E_(cell)^(@)-0.0591/2"log"([Zn^(2+)])/([H^(+)]^(2))`
`=0.760-0.0591/2"log"0.1/((10^(-7))^(2))=0.3759` volt