Question

Consider the cell, $H_{2_{(}g)}\left(1atm\right)|H_3{O}^{+}(pH=5.03)||{A{g}^{+}}_{(aq.)}\left(xM\right)|A{g}_{\left(s\right)}$
The measured EMF of the cell is $1.0V.$ What is the value of x?

$E_{A{g}^{+},Ag}^o=+0.8V.[T={25}^{o}C]$

• A. $2\times {10}^{-2}$
• B. $2\times {10}^{-3}$
• C. $1.5\times {10}^{-3}M$
• D. $1.5\times {10}^{-2}M$

Answer 1

The correct option is A $2\times {10}^{-2}$

$E_{cell}=E^0-\frac{0.059}{1}log\left[\frac{\left[H^{+}\right]}{\left[{Ag}^{+}\right]}\right]$

$E_{cell}=0.80-\frac{0.059}{1}log\left[\frac{\left[H^{+}\right]}{\left[{Ag}^{+}\right]}\right]$

$pH=5.03\Rightarrow \left[H^{+}\right]=9.33\times {10}^{-6}$

$1-0.80=\frac{-0.059}{1}log\left[\frac{9.33\times {10}^{-6}}{\left({Ag}^{+}\right)}\right]$

$\frac{-0.2}{0.059}=log\left[\frac{9.33\times {10}^{-6}}{\left[{Ag}^{+}\right]}\right]$

$4.075\times {10}^{-4}=\frac{9.33\times {10}^{-6}}{\left[{Ag}^{+}\right]}$

$\therefore \left[{Ag}^{+}\right]=2\times {10}^{-2}$