Chemistry - Electrochemistry Question with Solution | TestHub

Question Explanation:

calculate $\left[{\mathrm{H}}^{+}\right]$using Nernst equation for a coupled cell.

Concept:

Nernst Equation and Cell Potential.

Solution:

Step 1: Nernst for Fe³⁺/Fe²⁺

${\mathrm{E}}_{{\mathrm{Fe}}^{}}^{}=E^0+0.0591\frac{F{e}^{3+}}{F{e}^{2+}}$

$0.77+0.059log\frac{0.01}{0.02}$

$0.77+0.0591{\log }_{}0.5$

$log\left(0.5\right)=-0.301$

$${\mathrm{E}}_{{\mathrm{Fe}}^{}}^{}=0.77-0.0178=0.7522V$$

Step 2: Nernst for Cu²⁺

$$\begin{gathered} \text{displaylines}{\mathrm{E}}_{Cu}=0.34+\frac{0.0591}{2}log0.01 \\ =0.34-0.1182=02218V \end{gathered}$$

Step 3: Hydrogen electrode EMF:

${\mathrm{E}}_{H^{}}^{}=0-0.0591\left[H^{+}\right]$

Step 4: EMF of the cell: Hydrogen is anode (loses electrons)

$$\begin{gathered} \text{displaylines}{\mathrm{E}}_{cell}=E_{Cu}-E_H=0.36 \\ 0.36=0.2218-\left(-0.0591{\log }_{}\left[H^{+}\right]\right) \end{gathered}$$

${\log }_{}\left[{\mathrm{H}}^{+}\right]=\frac{0.36-0.2218}{0.0591}=2.34$

$$\left[{\mathrm{H}}^{+}\right]=10^{2.34}=0.22M$$