Chemistry - Electrochemistry Question with Solution | TestHub

Question Explanation: Calculating the concentration of ${\mathrm{H}}^{+}$ using the Nernst Equation.

Concept: Nernst Equation and Galvanic Cells.

Solution:

The cell is represented as $\mathrm{Pt}|{\mathrm{H}}_2|{\mathrm{H}}^{+}{\mathrm{Cu}}^{2+}|\mathrm{Cu}$.

For the hydrogen electrode:

${\mathrm{E}}_{\mathrm{H}}=-0.059\log [{\mathrm{H}}^{+}]$

For the copper electrode:

${\mathrm{E}}_{\mathrm{Cu}}=0.34+\frac{0.059}{2}\log [{\mathrm{Cu}}^{2+}]$

Given: $[{\mathrm{Cu}}^{2+}]=0.01=10^{-2}$

The cell potential for copper is:

${\mathrm{E}}_{\mathrm{Cu}}=0.34+0.0295\log (10^{-2})=0.34-0.059=0.281\mathrm{V}$

The overall cell potential is:

${\mathrm{E}}_{\text{cell }}={\mathrm{E}}_{\mathrm{Cu}}-{\mathrm{E}}_{\mathrm{H}}$

Substituting the given cell potential and the expressions for ${\mathrm{E}}_{\mathrm{Cu}}$ and ${\mathrm{E}}_{\mathrm{H}}$:

$0.10=0.281-\left(-0.059\log [{\mathrm{H}}^{+}]\right)$

$0.10=0.281+0.059\log [{\mathrm{H}}^{+}]$

$0.10-0.281=0.059\log [{\mathrm{H}}^{+}]$

$-0.181=0.059\log [{\mathrm{H}}^{+}]$

$\log [{\mathrm{H}}^{+}]=-3.07$

$[{\mathrm{H}}^{+}]\approx 10^{-3}=0.001\mathrm{M}$

Final answer: option (A)