Question

Calculate the voltage E in V, of the cell
$Ag\left(s\right)AgI{O}_3\left(s\right)\left|A{g}^{+}\right(xM),HI{O}_3(0.300M\left)\right|\left|Z{n}^{2+}\right(0.175M\left)\right|Zn\left(s\right)$
if $K_{SP}=3.02\times {10}^{-8}$ for $AgI{O}_3\left(s\right)$ and $K_a=0.162$ for $HI{O}_3,E^{\circ }\left(Z{n}^{2+}/Zn\right)=-0.76V,E^{\circ }(Ag/A{g}^{+}=-0.8V)$

• A. $0.45$
• B. $0.32$
• C. $1.88$
• D. $1.45$

Answer 1

The correct option is A $0.45$

$E_{cell}^0=E_R^0-E_L^0=-0.76-(-0.8)=0.04V$

	$HI{O}_3$	$H^{+}$	$I{O}_3^{-}$
Initial concentration (M)	0.300	0	0
Final concentration (M)	$0.300-x$	$x$	$x$

The equilibrium constant expression is
$K=\frac{\left[H^{+}\right]\left[I{O}_3^{-}\right]}{\left[HI{O}_3\right]}$
$0.162=\frac{x\times x}{0.3-x}$
$6.173x^2+x-0.3=0$
This is quadratic equation with solutions -0.31 and 0.154. As concentration cannot be negative, the value -0.31 is discarded.
$\left[I{O}_3^{-}\right]=x=0.154M$
$K_{sp}=\left[A{g}^{+}\right]\left[I{O}_3^{-}\right]$
$3.02\times {10}^{-8}=\left[A{g}^{+}\right]\times 0.154$
$\left[A{g}^{+}\right]=19.6\times {10}^{-8}=1.96\times {10}^{-7}M$
$Z{n}^{2+}+2Ag\to Zn+2A{g}^{+}$
$E_{cell}=E_{cell}^0-\frac{0.0592}{n}log\frac{[A{g}^{+}{]}^2}{\left[Z{n}^{2+}\right]}$
$E_{cell}=0.04-\frac{0.0592}{2}log\frac{(1.96\times {10}^{-7}{)}^2}{0.175}$
$E_{cell}=0.04+0.375=0.415V$