Question

Let's help Electra calculate the EMF of the cell $Zn\left(s\right)|Z{n}^{2+}\left(0.024M\right)||Z{n}^{2+}\left(2.4M\right)|Zn\left(s\right)$ at ${25}^{0}C$.
Round off the answer to 2 decimal places.

• A. 0.07 V
• B. 0.06 V
• C. 0.08 V
• D. 0.1 V

Answer 1

The correct option is B 0.06 V

The most important thing to note here is that at standard conditions. $E^{\circ }=0$
The half - cell reactions will be as follows:
$Z{n}^{2+}\left(2.4M\right)+2e^{-}\to Zn$ Reduction
$Zn\to Z{n}^{2+}\left(0.024M\right)+2e^{-}$ Oxidation

The Nernst equation for this reaction would be
$E_{cell}=E^{\circ }-\frac{2.303RT}{nF}log\left(\frac{\left[Z{n}^{2+}\right(0.024M\left)\right]\left[Zn\right(s\left)\right]}{\left[Z{n}^{2+}\right(2.4M\left)\right]\left[Zn\right(s\left)\right]}\right)$

Taking [Zn(s)] as unity, $E_0$ = 0 and plugging in n = 2, we end up with
$E_{cell}=0.0592V$

Rounding it off, we would get 0.06 V