Question

Electra has trouble finding the correct Nernst equation for the cell $Zn\left(s\right)|Z{n}^{2+}||C{u}^{2+}|Cu\left(s\right)at{25}^{\circ }C$
Can you help her?

• A. $E_{cell}=E^{\circ }-0.0296log\left(\frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]}\right)$
• B. $E_{cell}=E^{\circ }-0.0296log\left(\frac{\left[C{u}^{2+}\right]}{\left[Z{n}^{2+}\right]}\right)$
• C. $E_{cell}=E^{\circ }-0.0296log\left(\frac{Zn}{Cu}\right)$
• D. $E_{cell}=E^{\circ }-0.0296log\left(\frac{Cu}{Zn}\right)$

Answer 1

The correct option is A

$E_{cell}=E^{\circ }-0.0296log\left(\frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]}\right)$

Let us first recall the general form of Nernst equation which is
$E_{cell}=E^{\circ }-\frac{2.303RT}{nF}log\left(\frac{\left[C^c\right]\left[D^d\right]}{\left[A^a\right]\left[B^b\right]}\right)$

From the given cell reaction, we can write
Reduction on the right:$C{u}^{2+}\left(aq\right)+2e^{-}\to Cu\left(s\right)$
Oxidation on the left: $Zn\left(s\right)\to Z{n}^{2+}\left(aq\right)+2e^{-}$
Net reaction of cell is $Zn\left(s\right)+C{u}^{2+}\left(aq\right)\to Cu\left(s\right)+Z{n}^{2+}\left(aq\right)$

From here we see that, n = 2

We already know
$R=8.314J{K}^{-1}mo{l}^{-1}$
$T={25}^{\circ }C=298K$
$E_{cell}=E^{\circ }-\frac{2.303RT}{nF}log\left(\frac{\left[Z{n}^{2+}\right]\left[Cu\right(s\left)\right]}{\left[C{u}^{2+}\right]\left[Zn\right(s\left)\right]}\right)$

Don't forget that we consider the concentration of solids as unity.

If we put these into the above equation, we end up with
$E=E^{\circ }-0.0296log\left(\frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]}\right)$