Question

Consider the cell $\underset{1atm}{H_2\left(Pt\right)}|\underset{pH=5.00}{H_3{O}^{+}\left(aq\right)}||\underset{xM}{{\left(Ag\right)}^{+}}|Ag$
The measured $EMF$ of the cell is $1.0V$ . What is the value of $x?$ $E_{A{g}^{+},Ag}^0=+0.8V$ $\left[T=25^{\circ }C\right]$
$\left[\frac{2.303RT}{F}=0.06,log0.02=-\frac{5}{3}\right]$
Give your answer after multiplying 100 and round off to the nearest integer.

Answer 1

Answer: 2

Reaction for the cell is
$\frac{1}{2}{H}_2+A{g}^{+}\to H^{+}+Ag$
$E_{cell}=E_{cell}^o-\frac{0.06}{n}logQ1.0=0.80-\frac{0.06}{1}log\frac{10^{-5}}{x}log\frac{10^{-5}}{x}=-\frac{0.2}{0.06}-5-logx=-\frac{10}{3}logx=-5+\frac{10}{3}=-\frac{5}{3}x=0.02$