Question

For the cell: $$\mathrm{Cr}\left|\mathrm{Cr}^{3+} \| \mathrm{Co}^{2+}\right| \mathrm{Co}$$ \(E^{\circ}\) is \(0.46 \mathrm{~V}\). The same cell was prepared in the laboratory at standard conditions. The voltage obtained was \(0.40 \mathrm{~V}\). A possible explanation for the difference is (a) the surface area of the chromium electrode was smaller than the cobalt electrode. (b) the mass of the chromium electrode was larger than the mass of the cobalt electrode. (c) the concentration of \(\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}\) solution used was less than \(1.0 \mathrm{M}\) (d) the concentration of \(\mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{2}\) solution used was less than \(1.0 \mathrm{M}\) (e) the volume of \(\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}\) solution used was larger than the volume of \(\mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{2}\) solution used.

Short answer

The most likely explanation for the difference in the measured cell potential (0.40 V) compared to the standard cell potential (0.46 V) is (c) the concentration of Co(NO₃)₂ solution used was less than 1.0 M. This would affect the concentration of Co²⁺ ions and, as a result, lead to a lower cell potential according to the Nernst equation.

Step-by-step solution

The Nernst equation is used to find the non-standard cell potential (\(E\)) based on the standard cell potential (\(E^\circ\)) and the concentrations of the ions in the solutions. The equation is given by: $$E = E^\circ - \frac{0.0592}{n} \log Q$$ where \(n\) is the number of electrons transferred in the redox reaction, and \(Q\) is the reaction quotient, defined as the ratio of the product of the concentrations of the products to the product of the concentrations of the reactants, each raised to their respective stoichiometric coefficients. #Step 2: Identify the half-reactions and overall redox reaction#

We have a Cr|Cr³⁺||Co²⁺|Co electrochemical cell. The half reactions occurring at the electrodes are: Cathode (reduction): $$\mathrm{Co^{2+}} + 2e^− → \mathrm{Co}$$ Anode (oxidation): $$\mathrm{Cr} → \mathrm{Cr^{3+}} + 3e^−$$ The overall redox reaction is: $$\mathrm{Cr} + 2\mathrm{Co^{2+}} → \mathrm{Cr^{3+}} + 2\mathrm{Co}$$ #Step 3: Analyze the given options based on the Nernst Equation#

Referring to the Nernst equation and the overall redox reaction, the only factor that would affect the cell potential is a change in ion concentrations. We can now analyze each option to determine if any could explain the deviation from the standard cell potential: (a) A difference in electrode surface area would not affect cell potential. (b) A difference in electrode mass would not affect cell potential. (c) If the concentration of \(\mathrm{Co(NO_3)_2}\) is less than 1.0 M, that would affect the concentration of \(\mathrm{Co^{2+}}\) ions, which in turn will affect the cell potential. (d) If the concentration of \(\mathrm{Cr(NO_3)_2}\) is less than 1.0 M, that would affect the concentration of \(\mathrm{Cr^{3+}}\) ions, which in turn would affect the cell potential. (e) Varying the volume of the solutions would not affect cell potential directly, since the Nernst equation is dependent on concentrations, not volumes. #Step 4: Choose the most likely explanation for the difference in cell potential#

Based on the analysis in Step 3, we can conclude that the difference in cell potential could be due to a change in ion concentrations. Thus, the most likely explanations are (c) the concentration of \(\mathrm{Co(NO_3)_2}\) solution used was less than 1.0 M or (d) the concentration of \(\mathrm{Cr(NO_3)_2}\) solution used was less than 1.0 M. However, considering the fact that the experimental voltage is lower than the standard voltage, the most likely explanation would be that the concentration of \(\mathrm{Co(NO_3)_2}\) solution used was less than 1.0 M because a lower concentration of \(\mathrm{Co^{2+}}\) ions would lead to a lower cell potential according to the Nernst equation. So, the correct answer is (c).