Question

An ${\mathbf{\text{I}}}_{\mathbf{2}}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{\mid }{\mathbf{\text{I}}}^{\mathbf{-}}\mathbf{(}\mathbf{1}\mathbf{.00}\mathbf{\text{M}}\mathbf{)}$half-cell is connected to an${\mathbf{\text{H}}}_{\mathbf{3}}{\mathbf{\text{O}}}^{\mathbf{+}}\mathbf{\mid }{\mathbf{\text{H}}}_{\mathbf{2}}$ (1 atm) half-cell in which the concentration of the hydronium ion is unknown. The measured cell potential is$\mathbf{0}\mathbf{.841}\mathbf{}\mathbf{\text{V}}$ , and the${\mathbf{\text{I}}}_{\mathbf{2}}\mathbf{\mid }{\mathbf{\text{I}}}^{\mathbf{-}}$ half-cell is the cathode. What is the$\mathbf{\text{pH}}$ in the${\mathbf{\text{H}}}_{\mathbf{3}}{\mathbf{\text{O}}}^{\mathbf{+}}\mathbf{\mid }{\mathbf{\text{H}}}_{\mathbf{2}}$ half-cell?

Short answer

The $\text{pH}$ in the ${\text{H}}_3{\text{O}}^{+}\mid {\text{H}}_2$ half-cell is $5.17$

Step-by-step solution

Formula used

Nernst equation:

$E=E^{°}-\frac{0.0592\text{V}}{n}{\log }_{10}Q$

The electrode potential of cell:

$E^{°}=E_{\text{cathode}}^{°}-E_{\text{anode}}^{°}$

Calculate the electrode potential of the cell

Electrode potential of cell is

$\begin{array}{l}{E}_{\text{cathode}}^{°}=0.5355\text{V}\\ E_{\text{anode}}^{°}=0.000\\ E^{°}=0.5355\text{V}-0.000\\ =0.5355\text{V}\end{array}$

Put the values in the Nernst equation

Values are:

$\begin{array}{l}E=0.841V\\ E^0=0.5355V\end{array}$

$n=2$

$pH=-\log \left[{\text{H}}_3{\text{O}}^{+}\right]$

Put in Nernst equation

$\begin{array}{l}0.841\text{V}=0.5355\text{V}-\frac{0.0592\text{V}}{2}{\log }_{10}\frac{{\left[{\text{I}}^{-}\right]}^2{\left[{\text{H}}_3{\text{O}}^{+}\right]}^2}{P_{{\text{H}}_2}}\\ 0.841\text{V}=0.5355\text{V}-\frac{0.0592\text{V}}{2}{\log }_{10}\frac{{\left[1.0\right]}^2{\left[{\text{H}}_3{\text{O}}^{+}\right]}^2}{1}\\ 0.841\text{V}=0.5355\text{V}-\frac{0.0592\text{V}}{2}{\log }_{10}{\left[{\text{H}}_3{\text{O}}^{+}\right]}^2\end{array}$