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 grams
B. 4 grams
C. 10 grams
D. 2 grams

Answer

A

Explanation

`{:("Anode",Zn^(R)Zn^(2+)+2e^(-)),("Cathode",2H^++2e^(-) to H_2):}`, Cell:,`Zn+2H^+ hArr H_2+Zn^(2+), E_("cell")^(@)=0-(-0.76)=0.76 V`
`therefore 0.701=0.76-0.059/2 "log" (0.01xx1)/([H^+])^2`
`therefore [H^+]=10^(-2)` M
`:.` NaOH required is 0.01 mole =0.4 grams