You are given the followin cell at `298 K, Zn|(Zn^(+ +) ._((aq.))),(0.01M)||(HCl_((aq.))),(1.0lit)||(H_(2)(g)),(1.0atm)|Pt` with `E_(cell)=0.701` and `E_(Zn^(2+)//Zn)^(0)=-0.76V`. Which of the following amounts of `NaOH (` equivalent weight `=40 )` will just make the `pH` of cathodic compartment to be equal to `7.0 :`

Options

A. `0.4 g`
B. `4g`
C. `10g`
D. `2g`

Answer

A

Explanation

`{:("Anode",,ZnrarrZn^(2+)+2e^(-)),("Cathode",,2H^(+)+2e^(-)rarrH_(2)),("Cell:",,Zn+2H^(+)hArrH_(2)+Zn^(2+)" "E_(cell)^(@)=0-(-0.76)=0.76V):}`
`:." "0.701=0.76-(0.059)/(2)log.(0.01xx1)/([H^(+)]^(2))`
`:." "[H^(+)]=10^(-2)M`
`:. " "NaOH` required is `0.01` mole `=0.4 ` grams.