Question

Write the Nernst equation for the cell reaction in the Daniel cell. How will the $E_{\text{cell}}$ be affected when the concentration of $Z{n}^{2+}$ ions is increased?

Answer 1

Explaining the representation of Daniel cells:

Daniel Cell is represented as:

• Daniel cell involves the redox reaction
• The oxidation takes place at the anode while the reduction takes place at the cathode.

Cell Reaction:
$Zn\left(s\right)|Z{n}^{2+}\left(aq\right)||C{u}^{2+}\left(aq\right)|Cu\left(s\right)$

Writing the reaction of half a cell

Anode - Oxidation half - cell:

$Zn\left(s\right)\to Z{n}^{2+}\left(aq\right)+2e^{-}$

Cathode - Reduction half - cell:

$C{u}^{2+}\left(aq\right)+2e^{-}\to Cu\left(s\right)$

Overall reaction:

$Zn\left(s\right)+C{u}^{2+}\left(aq\right)\to Z{n}^{2+}\left(aq\right)+Cu\left(s\right)$

Writing the Nernst equation for the above Daniel cell reaction

$E_{\text{cell}}=E_{\text{cell}}^0-\frac{RT}{nF}\log \frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]}$

Where,

$E_{\text{cell}}=\text{Electric potential at any given concentration}$

$E_{\text{cell}}^0=\text{Standard electrode potential}$

$R=\text{Universal gas constant}=8.314J{K}^{-1}mo{l}^{-1}$

$F=\text{Faraday Constant}=96500Cmo{l}^{-1}$

$T=\text{Temperature in Kelvin}$

$\left[Z{n}^{2+}\right]$ and $\left[C{u}^{2+}\right]$ are the concentration of $Z{n}^{2+}$ and $C{u}^{2+}$ ions in the solution.

When we substitute the value for $R,F,T\left(298K\right)$, We can reduce the equation to:

$E_{\text{cell}}=E_{\text{cell}}^0-\frac{0.059}{2}\log \frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]}$

$E_{\text{cell}}$ decreases when the concentration $\left[Z{n}^{2+}\right]$ increases.