Question

By using the half-cell potentials in Appendix E, calculate the equilibrium constant at ${25}^{°}\text{C}$for the reaction in problem 33. Dichromate ion$\left({\text{Cr}}_2{\text{O}}_7^{2-}\right)$ is orange, and ${\text{Cr}}^{3+}$is light green in aqueous solution. If$2.00\text{L}$ of$1.00\text{MA}$ ${\text{HClO}}_2$solution is added to $2.00\text{L}$of $0.50\text{MCr}{\left({\text{NO}}_3\right)}_3$solution, what colour will the resulting solution have?

Short answer

Colour of the resulting solution where $K=3\times {10}^{31}$ is orange.

Step-by-step solution

Formula used

Equilibrium constant:

${\log }_{10}K=\frac{\text{n}}{0.059\text{V}}\times {\text{E}}^{°}$

The electrode potential of the cell:

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

Write Anode-cathode half-reactions

Cathode half-cell reaction:

3HClO_2_ $\left(aq\right)+6{\text{H}}_3{\text{O}}_{\left(aq\right)}^{+}+6{\text{e}}^{-}\to 3{\text{HClO}}_{\left(aq\right)}+9{\text{H}}_2{\text{O}}_{\left(l\right)}$

${\text{E}}^{°}=-1.33\text{V}$

Anode half-cell reaction:

$\begin{array}{l}2{\text{Cr}}^{3+}\left(aq\right)+21{\text{H}}_2{\text{O}}_{\left(l\right)}\to {\text{Cr}}_2{\text{O}}_7^{2-}\left(aq\right)+14{\text{H}}_3{\text{O}}^{+}\left(aq\right)6{\text{e}}^{-}\\ {\text{E}}^{°}=1.64\text{V}\end{array}$

Calculate the standard cell potential

The standard cell potential is

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

Calculate equilibrium constant

Equilibrium constant is

${\log }_{10}K=\frac{\text{n}}{0.059\text{V}}\times {\text{E}}^{°}$

Put $n=6$and$E^0=0.31V$

$\begin{array}{l}{\log }_{10}K=\frac{6}{0.059\text{V}}\times 0.31\text{V}\\ {\log }_{10}K=31.42\end{array}$

$\begin{array}{l}K={10}^{31.42}\\ K=3\times {10}^{31}\end{array}$

Colour of resulting solution where $K=3\times {10}^{31}$ is orange.