Question

Mr.Nernst had an apprentice named Electra. She is hard-working but is not very good with Chemistry. She needs a lot of help from you to pass as an apprentice. Please help the damsel in distress!

One day Mr.Nernst told her to find the EMF of the following cell using experiments at ${25}^{\circ }C$
$Mg\left(s\right)+2A{g}^{+}\left(0.0001M\right)\to M{g}^{2+}\left(0.130M\right)+2Ag\left(s\right)$

Electra reported the value to be +2.6V. Mr.Nernst told her to recheck the value using his theory, knowing Standard EMF = 3.17 V.
What was the right value? Round off the answer to nearest integer.

Type your answer here. ___

Answer 1

Let's help Electra together.

First, write down the Nernst equation's general form for this reaction. Observe that I haven't included the concentrations of Mg(s) and Ag(s) since these are taken as unity.
$E_{cell}=E^{\circ }-\frac{2.303RT}{nF}log\left(\frac{\left[M{g}^{2+}\right]}{[A{g}^{2+}{]}^2}\right)$

$E^{\circ }=3.17V$ as given in the question.
$R=8.314J{K}^{-1}mo{l}^{-1}$
$T={25}^{\circ }C=298K$
Also, $\left[M{g}^{2+}\right]=0.310M$
$\left[A{g}^{+}\right]=0.0001Mor{10}^{-4}M$

To find n, we need to write down the half-reactions of the cell and balance the electrons. Let's do that. Let's take the equation given as reference. We can see that,
$Mg\to M{g}^{2+}+2e^{-}$
$2A{g}^{+}+2e^{-}\to 2Ag$

So, we see that the reactions is balanced and n = 2
By plugging in all these values, we get
$E_{cell}=E^{\circ }-\frac{2.303RT}{nF}log\left(\frac{\left[M{g}^{2+}\right]}{[A{g}^{2+}{]}^2}\right)$
$E_{cell}=E_0-\frac{2.303\ast 8.314\ast 298}{n\ast 96500}log\left(\frac{0.13}{{10}^{-3}}\right)$

If we calculate the value, we'll get $E_{cell}=2.96V$. We can round that off to 3 V