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Question: Given, \(E_{C{{r}^{2+}}/Cr}^{\circ }=-0.72V,E_{F{{e}^{2+}}/Fe}^{\circ } = -0.42V.\) The potential ...

Given, ECr2+/Cr∘=−0.72V,EFe2+/Fe∘=−0.42V.E_{C{{r}^{2+}}/Cr}^{\circ }=-0.72V,E_{F{{e}^{2+}}/Fe}^{\circ } = -0.42V.
The potential for the cell Cr∣Cr3+(0.1M)∥Fe2+(0.01M)∣Fe\left. Cr \right|C{{r}^{3+}}(0.1M)\parallel F{{e}^{2+}}\left. (0.01M) \right|Fe is:
A. 0.26 V
B. 0.399 V
C. -0.399 V
D. -0.26 V

Explanation

Solution

In this process the transfer of electrons that changes the oxidation state of the ions or molecule. There are basically two types of cells that are electrolytic and voltaic cells. Here, Nernst equation is the principle behind this numerical.

Complete Solution :
- We can see that in the cell diagram, the first part denotes the anodic part and the last part denotes the cathodic part.
Cr∣Cr3+(0.1M)∥Fe2+(0.01M)∣Fe\left. Cr \right|C{{r}^{3+}}(0.1M)\parallel F{{e}^{2+}}\left. (0.01M) \right|Fe
- In the cell diagram we can see that chromium is oxidised to Cr3+C{{r}^{3+}} and Fe2+F{{e}^{2+}} get reduced to Fe.
- We can write the oxidation half-cell as:
Cr→Cr3++3e−]×2Cr\to C{{r}^{3+}}+3{{e}^{-}}]\times 2
- We can write the reduction half-cell as:
Fe2++2e−→Fe]×3F{{e}^{2+}}+2{{e}^{-}}\to Fe]\times 3

- Now, the net cell reaction can be written as:
2Cr+Fe2+→2Cr3++3Fe,n=62Cr+F{{e}^{2+}}\to 2C{{r}^{3+}}+3Fe,n=6
- The is given by the formula:
Ecell∘=Eoxi∘+Ered∘{{E}_{cell}}^{\circ }={{E}_{oxi}}^{\circ }+{{E}_{red}}^{\circ }
- By substituting the values in the above equation we get:
=0.72−0.42=0.30V=0.72 - 0.42 = 0.30V

Now, we will find the value of Ecell{{E}_{cell}} as:

& {{E}_{cell}}=E_{cell}^{\circ }-\dfrac{0.059}{n}\log \dfrac{{{\left[ C{{r}^{3+}} \right]}^{2}}}{{{\left[ F{{e}^{2+}} \right]}^{3}}} \\\ & = 0.30-\dfrac{0.059}{6}\log \dfrac{{{\left[ 0.1 \right]}^{2}}}{{{\left[ 0.01 \right]}^{3}}} \\\ & = 0.30-\dfrac{0.059}{6}\log \dfrac{{{10}^{-2}}}{{{10}^{-6}}} \\\ & = 0.30-\dfrac{0.059}{6}\log {{10}^{4}} \\\ & {{E}_{cell}}=0.2606V \\\ \end{aligned}$$ \- Hence, we can conclude that the correct option is (a), that is the potential for the cell $\left. Cr \right|C{{r}^{3+}}(0.1M)\parallel F{{e}^{2+}}\left. (0.01M) \right|Fe$ is 0.26 V. **So, the correct answer is “Option B”.** **Note:** \- It is found that Gibbs’ free energy for a cell can be calculated by using Nernst equation. It is found that at equilibrium forward and reverse reactions occur at equal rates. Hence, the cell potential is basically zero volts. \- This equation allows us to calculate the cell potential when the two cells are not in 1 M concentration. We can calculate the equilibrium constant from the cell potential.