Question

Consider the cell $\begin{array}{c}{H}_2\left(Pt\right)\\ 1atm\end{array}\left|\begin{array}{c}{H}_3{O}^{+}\\ pH=5.5\end{array}\right|\left|\begin{array}{c}{Ag}^{+}\\ xM\end{array}\right|Ag$. The measured EMF of the cell is $1.023V$. What is the value of $x$?
$E_{{Ag}^{+}.Ag}^0=+0.799V(T={25}^{o}C)$

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

Answer 1

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

Cathode: $A{g}^{+}+e^{-}⟶Ag$ $E_{Ag+/Ag}^o=0.799V$

Anode: $1/2H_2⟶H^{+}+e^{-}$ $E_{H_2/H^{+}}^o=OV$

$A{g}^{+}+1/2H_2⟶Ag+H^{+}$

$E_{cell}^o=0.799-0=0.799V$

Using Nernst Equation, at ${25}^{o}C$

$E_{cell}=E_{cell}^o-\frac{0.0591}{1}\log \frac{\left[H^{+}\right]}{\left[A{g}^{+}\right]}$

as $pH=5.5=-\log \left[H^{+}\right]$

thus $\left[H^{+}\right]={10}^{-5.5}$

$\Rightarrow 1.023=0.799-\frac{0.0591}{1}\log \frac{\left[{10}^{-5.5}\right]}{\left[x\right]}$

$\Rightarrow 0.224=5.5\times 0.0591+0.0591\log \left[x\right]$

On solving, we get

$\Rightarrow \left[x\right]=2\times {10}^{-2}M$