Question

Question: A half-cell has a graphite electrode immersed in an acidic solution (pH0) of Mn²⁺(concentration (1.00M))in contact with solid MnO₂. A second half-cell has an acidic solution (pH0) of H₂O₂(concentration (1.00M)in contact with a platinum electrode past which gaseous oxygen at a pressure of 1 atmis bubbled. The two half-cells are connected to form a galvanic cell.

(a) Referring to Appendix E, write balanced chemical equations for the half-reactions at the anode and the cathode and for the overall cell reaction.

(b) Calculate the cell potential.

Short answer

Answer

a) The balanced equation is

${\text{MnO}}_2\left(s\right)+2{\text{H}}_3{\text{O}}^{+}\left(aq\right)+{\text{H}}_2{\text{O}}_2\left(aq\right)\to {\text{Mn}}^{2+}\left(aq\right)+{\text{O}}_2\left(g\right)+4{\text{H}}_2\text{O}\left(l\right)$

b) The cell potential is

Step-by-step solution

The cell potential and half-cell potential.

The cell potential is the measurement of the potential difference between two half cells in an electrochemical cell.

It can be calculated by the following equation.

$\left(E_{\text{cell}}^0\right)\mathbf{=}{\mathit{E}}_{\mathbf{\text{caltiode}}}^{\mathbf{0}}\mathbf{-}{\mathit{E}}_{\mathbf{\text{anode}}}^{\mathbf{0}}$

Half-cell potential refers to the potential at the electrode of each half cell in an electrochemical cell. In an electrochemical cell, the total potential is the total potential calculated from the potentials of two half cells.

${\mathit{E}}_{\mathbf{\text{cell}}}\mathbf{=}{\mathit{E}}_{\mathbf{\text{cell}}}^{\mathbf{0}}\mathbf{-}\frac{\mathbf{0}\mathbf{.}\mathbf{0592}\mathbf{\text{V}}}{\mathbf{n}}\mathit{l}\mathit{o}{\mathit{g}}_{\mathbf{10}}{\mathit{Q}}_{\mathbf{\text{cell}}}$

The overall reaction.

a)

From the statement given below, is more positive, so will be reduced at cathode.

$$\begin{gathered} E^0\left({\text{MnO}}_2\mid {\text{Mn}}^{2+}\right)=1.208\text{V} \\ {E}^0\left({\text{H}}_2{\text{O}}_2\mid {\text{O}}_2\right)=0.682\text{V} \end{gathered}$$

The cathode half-cell reaction

${\text{MnO}}_2\left(s\right)+4{\text{H}}_3{\text{O}}^{+}\left(aq\right)+2e^{-}\to {\text{Mn}}^{2+}\left(aq\right)+6{\text{H}}_2\text{O}\left(l\right)$

The anode half-cell reaction${\text{H}}_2{\text{O}}_2\left(aq\right)+2{\text{H}}_2\text{O}\left(l\right)\to {\text{O}}_2\left(g\right)+2{\text{H}}_3{\text{O}}^{+}\left(aq\right)+2e^{-}$

The overall reaction

${\text{MnO}}_2\left(s\right)+2{\text{H}}_3{\text{O}}^{+}\left(aq\right)+{\text{H}}_2{\text{O}}_2\left(aq\right)\to {\text{Mn}}^{2+}\left(aq\right)+{\text{O}}_2\left(g\right)+4{\text{H}}_2\text{O}\left(l\right)$

Find the cathode half-cell potential.

b)

Cathode half-cell potential is given by,

$E_{\text{hc}}=E_{\text{hc}}^0-\frac{0.0592\text{V}}{n}{\log }_{10}{Q}_{\text{hc}}$

The cathode half-cell potential

$$\begin{gathered} E\left({\text{MnO}}_2\mid {\text{Mn}}^{2+}\right)=1.208\text{V}-\frac{0.0592\text{V}}{2}{\log }_{10}\frac{\left[{\text{Mn}}^{2+}\right]}{{\left[{\text{H}}_3{\text{O}}^{+}\right]}^4} \\ E\left({\text{MnO}}_2\mid {\text{Mn}}^{2+}\right)=1.208\text{V}-\frac{0.0592\text{V}}{2}{\log }_{10}\frac{1.0}{{(1.0)}^4} \\ E\left({\text{MnO}}_2\mid {\text{Mn}}^{2+}\right)=1.208\text{V}-\frac{0.0592\text{V}}{2}\times 0 \\ E\left({\text{MnO}}_2\mid {\text{Mn}}^{2+}\right)=1.208\text{V} \end{gathered}$$

The anode half-cell potential

The anode half-cell potential

$$\begin{gathered} E\left({\text{H}}_2{\text{O}}_2\mid {\text{O}}_2\right)=0.682\text{V}-\frac{0.0592\text{V}}{2}{\log }_{10}\frac{P_{{\text{O}}_2}{\left[{\text{H}}_3{\text{O}}^{+}\right]}^2}{\left[{\text{H}}_2{\text{O}}_2\right]} \\ E\left({\text{H}}_2{\text{O}}_2\mid {\text{O}}_2\right)=0.682\text{V}-\frac{0.0592\text{V}}{2}{\log }_{10}\frac{1\times {(1.0)}^2}{1.0} \\ E\left({\text{H}}_2{\text{O}}_2\mid {\text{O}}_2\right)=0.682\text{V}-\frac{0.0592\text{V}}{2}\times 0 \\ E\left({\text{H}}_2{\text{O}}_2\mid {\text{O}}_2\right)=0.628\text{V} \end{gathered}$$

Calculate the cell potential   

Cell potential

$$\begin{gathered} \left(E_{\text{cell}}^0\right)=E_{\text{caltiode}}^0-E_{\text{anode}}^0 \\ =1.208\text{V}-0.682\text{V} \\ =0.526\text{V} \end{gathered}$$