Question

${\mathbf{\text{ACu}}}^{\mathbf{2}\mathbf{+}}\mathbf{(}\mathbf{1}\mathbf{.00}\mathbf{\text{M}}\mathbf{)}\mathbf{\mid }\mathbf{\text{Cu}}$half-cell is connected to ${\mathbf{\text{Br}}}_{\mathbf{2}}\mathbf{\left(}\mathcal{l}\mathbf{\right)}\mathbf{\mid }{\mathbf{\text{Br}}}^{\mathbf{-}}$half-cell in which the concentration of bromide ion is unknown. The measured cell potentialis $\mathbf{0}\mathbf{.963}\mathbf{}\mathbf{\text{V}}$,and the ${\mathbf{\text{Cu}}}^{\mathbf{2}\mathbf{+}}\mathbf{\mid }\mathbf{\text{Cu}}$half-cell is the anode. What is the bromide ion concentration in the${\mathbf{\text{Br}}}_{\mathbf{2}}\mathbf{\left(}\mathcal{l}\mathbf{\right)}\mathbf{\mid }{\mathbf{\text{Br}}}^{\mathbf{-}}$ half-cell?

Short answer

The bromide ion concentration in the${\text{Br}}_2\left(\mathcal{l}\right)\mid {\text{Br}}^{-}$ half-cell is$6.725\times {10}^{-5}\text{M}.$

Step-by-step solution

Formula used

Nernst equation:

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

The electrode potential of the cell:

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

Calculate the electrode potential of the cell 

Electrode potential of the cell is

$\begin{array}{l}{E^0}_{anode}=0.340V\\ {E^0}_{cathode}=1.056V\end{array}$

$\begin{array}{l}{E^0}_{cell}=1.056-0.340\\ =0.716V\end{array}$

Put the values in the Nernst equation

Values are:

$\begin{array}{l}E=0.963V\\ {E^0}_{cell}=0.716V\end{array}$

$n=2$

Put in Nernst equation

$\begin{array}{l}{E}_{\text{cell}}=E_{\text{cell}}^o-\frac{0.0592\text{V}}{n}{\log }_{10}\frac{{\left[{\text{Br}}^{-}\right]}^2}{\left[{\text{Cu}}^{+2}\right]}\\ 0.963\text{V}=0.716\text{V}-\frac{0.0592\text{V}}{2}{\log }_{10}\frac{{\left[{\text{Br}}^{-}\right]}^2}{1.00\text{M}}\\ 0.963\text{V}=0.716\text{V}-\frac{0.0592\text{V}}{2}2{\log }_{10}\frac{\left[{\text{Br}}^{-}\right]}{1.00M}\end{array}$

$\begin{array}{l}\frac{0.247\text{V}}{-0.0592\text{V}}={\log }_{10}\frac{\left[{\text{Br}}^{-}\right]}{1.00M}\\ -4.1723={\log }_{10}\frac{\left[{\text{Br}}^{-2}\right]}{1.00M}\\ \left[{\text{Br}}^{-}\right]={10}^{-4.1723}\times 1.00M\\ \left[{\text{Br}}^{-}\right]=6.725\times {10}^{-5}\text{M}\end{array}$