Question

Let's help Electra calculate the EMF of the cell $Zn\left(s\right)|Z{n}^{2+}\left(0.024M\right)||Z{n}^{2+}\left(2.4M\right)|Zn\left(s\right)$. 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. $E^{\circ }=0$ because we would have the same concentration of $Z{n}^{2+}$ ions (unity) on both sides which is the only driving force in this case.

Also, flow of ions would be from concentrated solution to the dilute solution, which means that $Z{n}^{2+}$ ions from the concentrated solution would be reduced and ones from dilute solution would get oxidised.

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

The $Z{n}^{2+}$ ions try to move from the concentrated half cell to a dilute solution. That driving force gives rise to 0.0592 V.