Question

A galvanic cell is constructed in which the overall reaction is

${\mathbf{\text{Cr}}}_{\mathbf{2}}{\mathbf{\text{O}}}_{\mathbf{7}}^{\mathbf{2}\mathbf{-}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}\mathbf{14}{\mathbf{\text{H}}}_{\mathbf{2}}{\mathbf{\text{O}}}^{\mathbf{+}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}\mathbf{6}{\mathbf{\text{I}}}^{\mathbf{-}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{\to }\mathbf{2}{\mathbf{\text{Cr}}}^{\mathbf{3}\mathbf{+}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}\mathbf{3}{\mathbf{\text{I}}}_{\mathbf{2}}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{+}\mathbf{21}{\mathbf{\text{H}}}_{\mathbf{2}}\mathbf{\text{O}}\mathbf{\left(}\mathcal{l}\mathbf{\right)}$

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

b) At$\mathit{p}\mathit{H}\mathbf{=}\mathbf{0}$ ,with $\mathbf{\left[}{\mathbf{\text{Cr}}}_{\mathbf{2}}{\mathbf{\text{O}}}_{\mathbf{7}}^{\mathbf{2}\mathbf{-}}\mathbf{\right]}\mathbf{=}\mathbf{1}\mathbf{.5}\mathbf{\text{M}}$and,$\mathbf{\left[}{\mathbf{\text{I}}}^{\mathbf{-}}\mathbf{\right]}\mathbf{=}\mathbf{0}\mathbf{.40}\mathbf{\text{M}}$the cell potential is found to equal .$\mathbf{0}\mathbf{.87}\mathbf{}\mathbf{\text{V}}$ Calculate the concentration of ${\mathbf{\text{Cr}}}^{\mathbf{3}\mathbf{+}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}$in the cell.

Short answer

a)

$E^0=0.795V$

b)

the concentration of ${\text{Cr}}^{3+}\left(aq\right)$ in the cell is.$8.04\times {10}^{-4}M$

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 

a)

The electrode potential of the cell is

$\begin{array}{l}{E}_{\text{cathode}}^{°}=1.33\text{V}\\ E_{\text{anode}}^{°}=0.535V\\ E^{°}=1.33\text{V}-0.535V\\ =0.795\text{V}\end{array}$

Put the values in the Nernst equation

b)

Values are:

$\begin{array}{l}{E}_{cell}=0.87V\\ E^0=0.795V\end{array}$

$n=6$

$pH=0$

$\begin{array}{l}\left[{\text{H}}_3{\text{O}}^{+}\right]=1\\ \left[{\text{Cr}}_2{\text{O}}_7^{2-}\right]=1.5\text{M}\\ \left[I^{-}\right]=0.40\text{M}\end{array}$

Put in Nernst equation

$\begin{array}{l}0.87\text{V}=0.795\text{V}-\frac{0.0592\text{V}}{6}{\log }_{10}\frac{{\left[{\text{Cr}}^{3+}\right]}^2}{\left[1.5\right]{\left[0.40\right]}^6{\left[1\right]}^{14}}\\ 0.87\text{V}=0.795\text{V}-\frac{0.0592\text{V}}{6}{\log }_{10}\left[\frac{{\left[{\text{Cr}}^{3+}\right]}^2}{6.144\times {10}^{-3}}\right]\\ 0.87\text{V}=0.795\text{V}-\frac{0.0592\text{V}}{6}(2{\log }_{10}{\left[{\text{Cr}}^{3+}\right]}^2-{\log }_{10}(6.144\times {10}^{-3})\end{array}$

$\begin{array}{l}0.87\text{V}=0.795\text{V}-\frac{0.0592\text{V}}{6}(2{\log }_{10}{\left[{\text{Cr}}^{3+}\right]}^2-{\log }_{10}(6.144\times {10}^{-3}))\\ 0.87\text{V}=0.795\text{V}-\frac{0.0592\text{V}}{6}\left[2\log \left[{\text{Cr}}^{3+}\right]-(-2.21)\right]\\ 2\log \left[{\text{Cr}}^{3+}\right]+2.21=\frac{(0.795\text{V}-0.87\text{V})\times 6}{0.0592\text{V}}\\ 2\log \left[{\text{Cr}}^{3+}\right]+2.21=-7.60\end{array}$

$\begin{array}{l}2\log \left[{\text{Cr}}^{3+}\right]=-7.60-2.21\\ \log \left[{\text{Cr}}^{3+}\right]=\frac{-9.81}{2}=-4.905\\ \left[{\text{Cr}}^{3+}\right]={10}^{-4.905}=8.04\times {10}^{-4}\text{M}\end{array}$