Question

The following reaction occurs in an electrochemical cell:

$\mathbf{3}{\mathbf{\text{HClO}}}_{\mathbf{2}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{\text{Cr}}}^{\mathbf{3}\mathbf{+}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}\mathbf{12}{\mathbf{\text{H}}}_{\mathbf{2}}\mathbf{\text{O}}\mathbf{\left(}\mathcal{l}\mathbf{\right)}\mathbf{\to }\mathbf{3}\mathbf{\text{HClO}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}{\mathbf{\text{Cr}}}_{\mathbf{2}}{\mathbf{\text{O}}}_{\mathbf{7}}^{\mathbf{2}\mathbf{-}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}\mathbf{8}{\mathbf{\text{H}}}_{\mathbf{3}}{\mathbf{\text{O}}}^{\mathbf{+}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}$

a) Calculate $\mathit{\Delta }{\mathit{E}}^{\mathbf{0}}$for this cell.

b) At $\mathit{p}\mathit{H}\mathbf{=}\mathbf{0}$,with, $\mathbf{\left[}{\mathbf{\text{Cr}}}_{\mathbf{2}}{\mathbf{\text{O}}}_{\mathbf{7}}^{\mathbf{2}\mathbf{-}}\mathbf{\right]}\mathbf{=}\mathbf{0}\mathbf{.80}\mathbf{\text{M}}\mathbf{\left[}{\mathbf{\text{HClO}}}_{\mathbf{2}}\mathbf{\right]}\mathbf{=}\mathbf{0}\mathbf{.15}\mathbf{\text{M}}$ ,and$\mathbf{\left[}\mathbf{\text{HCIO}}\mathbf{\right]}\mathbf{=}\mathbf{0}\mathbf{.20}\mathbf{\text{M}}$ the cell potential is found tobe.$\mathbf{0}\mathbf{.15}\mathbf{}\mathbf{\text{V}}$Calculate the concentration of ${\mathbf{\text{Cr}}}^{\mathbf{3}\mathbf{+}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}$in the cell.

Short answer

a)

$E^0=0.31V$

b)

the concentration of ${\text{Cr}}^{3+}\left(aq\right)$ in the cell is.$1.0\times {10}^{-8}M$

Step-by-step solution

Formula used 

Nernst equation:

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

Electrode potential of cell:

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

Calculate the electrode potential of the cell

a)

The electrode potential of the cell is

$\begin{array}{l}{E}_{\text{cathode}}^{°}=1.64\text{V}\\ E_{\text{anode}}^{°}=1.33V\\ E^{°}=1.64\text{V}-1.33V\\ =0.31\text{V}\end{array}$

$\begin{array}{l}\left[{\text{H}}_3{\text{O}}^{+}\right]=1\\ \left[{\text{Cr}}_2{\text{O}}_7^{2-}\right]=0.80\text{M}\\ \left[{\text{HClO}}_2\right]=0.15\text{M}\\ \left[\text{HClO}\right]=0.20\text{M}\end{array}$

Put in Nernst equation

$\begin{array}{l}E=E^{°}-\frac{0.0592\text{V}}{n}{\log }_{10}\frac{\left[{\text{Cr}}_2{\text{O}}_7^2\right][{\text{HClO}}^3{\left[{\text{H}}_3{\text{O}}^{+}\right]}^8}{{\left[{\text{Cr}}^{3+}\right]}^2{\left[{\text{HClO}}_2\right]}^3}\\ 0.15\text{V}=0.31\text{V}-\frac{0.0592\text{V}}{6}{\log }_{10}\frac{\left[0.80\right]{\left[0.20\right]}^3{\left[1.0\right]}^8}{{\left[{\text{Cr}}^{3+}\right]}^2{\left[0.15\right]}^3}\\ 0.15\text{V}=0.31\text{V}-\frac{0.0592\text{V}}{6}{\log }_{10}\frac{1.9}{{\left[{\text{Cr}}^{3+}\right]}^2}\end{array}$

$\begin{array}{l}0.15\text{V}=0.31\text{V}-\frac{0.0592\text{V}}{6}({\log }_{10}1.9-2{\log }_{10}\left[{\text{Cr}}^{3+}\right])\\ 0.15\text{V}=0.31\text{V}-\frac{0.0592\text{V}}{6}(0.28-2{\log }_{10}\left[{\text{Cr}}^{3+}\right])\\ (0.28-2{\log }_{10}\left[{\text{Cr}}^{3+}\right])=\frac{(0.31\text{V}-0.15\text{V})\times 6}{0.0592\text{V}}\end{array}$

$\begin{array}{l}-2{\log }_{10}\left[{\text{Cr}}^{3+}\right]=16-0.28\\ {\log }_{10}\left[{\text{Cr}}^{3+}\right]=-\frac{15.72}{2}=-7.86\\ \left[{\text{Cr}}^{3+}\right]=1.0\times {10}^{-8}\text{M}\end{array}$