Question

Question:A galvanic cell is based on the following half-reactions:

$$\begin{gathered} {\mathbf{Fe}}^{\mathbf{+}\mathbf{2}}\mathbf{+}\mathbf{2}{\mathbf{e}}^{\mathbf{-}}\stackrel{}{\mathbf{\to }}\mathbf{Fe}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{E}}^{\mathbf{\circ }}\mathbf{=}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{440} \\ \mathbf{2}{\mathbf{H}}^{\mathbf{+}}\mathbf{+}\mathbf{2}{\mathbf{e}}^{\mathbf{-}}\stackrel{}{\mathbf{\to }}{\mathbf{H}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{E}}^{\mathbf{\circ }}\mathbf{=}\mathbf{0} \end{gathered}$$

In this cell, the iron compartment contains an iron electrodeand $\left[{\mathrm{Fe}}^{+2}\right]\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{atm}$, and the hydrogen compartment contains a platinum electrode,role="math" localid="1663906376319" ${\mathbf{pH}}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{atm}$and a weak acid HA at an initial concentration of role="math" localid="1663906339580" $\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{M}$.If the observed cell potential is $\mathbf{0}\mathbf{.}\mathbf{333}\mathbf{}\mathbf{V}$at ${\mathbf{25}}^{\mathbf{\circ }}\mathbf{C}$, calculate the Ka value for the weak acid HA atrole="math" localid="1663906523400" ${\mathbf{25}}^{\mathbf{\circ }}\mathbf{C}$.

Short answer

$K_a=2.43\times {10}^7$

Step-by-step solution

analyzing the given data

Given equations are;

$$\begin{gathered} {\mathrm{Fe}}^{+2}+2{\mathrm{e}}^{-}\stackrel{}{\to }\mathrm{Fe}\left(\mathrm{s}\right){\mathrm{E}}^{\circ }=-0.440 \\ 2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}\stackrel{}{\to }{\mathrm{H}}_2\left(\mathrm{g}\right){\mathrm{E}}^{\circ }=0 \end{gathered}$$

Flipping the first equation, we get,

$$\begin{gathered} {\mathrm{Fe}}^{+2}+2{\mathrm{e}}^{-}\stackrel{}{\to }\mathrm{Fe}\left(\mathrm{s}\right){\mathrm{E}}^{\circ }=-0.440 \\ {\mathrm{H}}_2\left(\mathrm{g}\right)\stackrel{}{\to }2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}{\mathrm{E}}^{\circ }=0 \end{gathered}$$

The final equation is,

$$\begin{gathered} {\mathrm{Fe}}^{+2}+{\mathrm{H}}_2\left(\mathrm{g}\right)\stackrel{}{\to }2{\mathrm{H}}^{+}+\mathrm{Fe}\left(\mathrm{s}\right) \\ {\mathrm{E}}_{\mathrm{cell}}^{\circ }={\mathrm{E}}_{\mathrm{cathode}}^{\circ }-{\mathrm{E}}_{\mathrm{anode}}^{\circ } \\ {\mathrm{E}}_{\mathrm{cell}}^{\circ }=0-\left(-0.440\right) \\ {\mathrm{E}}_{\mathrm{cell}}^{\circ }=0.440\mathrm{V} \end{gathered}$$

Calculating the value of ka

Using Nernst equation, we can calculate the concentration of ${\mathrm{H}}^{+}$ions.

localid="1663907701270" $$\begin{gathered} E_{cell}=E_{cell}^{\circ }-\frac{0.0591}{n}\log \frac{\left[F{e}^{+2}\right]}{{\left[H^{+}\right]}^2} \\ 0.333V=0.440V-\frac{0.0591}{2}\log \frac{1\times {10}^{-3}}{{\left[H^{+}\right]}^2} \\ 0.107=\frac{0.0591}{2}\log \frac{1\times {10}^{-3}}{{\left[H^{+}\right]}^2} \\ 3.82=\log \frac{1\times {10}^{-3}}{{\left[H^{+}\right]}^2} \\ 6668.48=\frac{1\times {10}^{-3}}{{\left[H^{+}\right]}^2} \\ {\left[H^{+}\right]}^2=14.9\times {10}^{-7} \\ \left[H^{+}\right]=4.89\times {10}^{-4}\mathrm{M} \end{gathered}$$

Since,$\left[H^{+}\right]=\left[A^{-}\right]$$\left[H^{+}\right]=\left[A^{-}\right]$

$K_a=\frac{{\left(4.89\times {10}^{-4}\right)}^2}{1.00}=2.3\times {10}^{-7}$