Problem 2 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

We have an oxidation-reduction system: \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}+\mathrm{e}^{-} \rightleftharpoons\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}\) \(E^{\circ}=+0.36 \mathrm{~V}\). The ratio of concentrations of oxidized and reduced from at which the potential of the system becomes \(0.24 \mathrm{~V}\), is [Given: \(2.303 R T / F=0.06\) ) (a) \(2: 1\) (b) \(1: 2\) (c) \(1: 20\) (d) \(1: 100\)

Step-by-Step Solution

Answer

The ratio of concentrations of oxidized to reduced form is approximately \(10^{2.03}\), which is closest to \(1: 100\), corresponding to option (d).

Step 1: Understand the Nernst Equation

The Nernst equation relates the concentration of chemical species to the potential of a reaction. For a reduction reaction, the Nernst equation at 25°C (298 K) is given by: \(E = E^{\text{o}} - \frac{0.059}{n} \log\frac{[oxidized]}{[reduced]}\), where \(E\) is the cell potential, \(E^{\text{o}}\) is the standard cell potential, \(n\) is the number of moles of electrons transferred, and the ratio \(\frac{[oxidized]}{[reduced]}\) represent the concentrations of the oxidized and reduced forms of the species.

Step 2: Calculate the Number of Moles of Electrons Transferred

From the given oxidation-reduction system equation, we can see that the number of moles of electrons transferred, \(n\), is 1, since there is a gain of one electron on the right side of the equation.

Step 3: Insert Known Values into Nernst Equation

Substitute the known values into the Nernst equation: \(0.24 = 0.36 - \frac{0.059}{1} \log\frac{[oxidized]}{[reduced]}\). Now solve for the concentration ratio.

Step 4: Solve for the Oxidized to Reduced Concentration Ratio

By rearranging the equation, we isolate the logarithm: \(0.059 \log\frac{[oxidized]}{[reduced]} = 0.36 - 0.24\). This simplifies to: \(\log\frac{[oxidized]}{[reduced]} = \frac{0.12}{0.059}\). Now calculate the value of the logarithm and the ratio.

Step 5: Calculate the Concentration Ratio

Calculate the value of the ratio \(\frac{[oxidized]}{[reduced]}\): \(\log\frac{[oxidized]}{[reduced]} = 2.03\), so \(\frac{[oxidized]}{[reduced]} = 10^{2.03}\). This gives us the ratio of the concentrations.

Step 6: Determine the Closest Ratio

The calculated ratio from the previous step should be compared to the options given to determine the closest answer.

Key Concepts

Oxidation-Reduction Reactions

The dance of electrons during chemical reactions is governed by a process called oxidation-reduction, or redox reactions. These are reactions where one substance loses electrons (oxidation) and another gains electrons (reduction). It's a balancing act—whenever a substance is oxidized, another is reduced. In the context of the educational problem provided, we have the redox couple \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}\) and \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}\text{.}\) Here, \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}\) loses an electron and gets oxidized, while \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}\) gains an electron and gets reduced.

Understanding redox reactions is pivotal because they underpin various biochemical processes and industrial applications like batteries and electroplating. Hence, this process directly connects to the utility of the Nernst Equation in evaluating the potential of an oxidation-reduction system.

Electrochemistry

Electrochemistry is the branch of chemistry that deals with the relationship between electricity and chemical change. In our problem, we delve into the heart of electrochemistry by looking at the potential, measured in volts, that arises from the redistribution of electrons during the redox reaction. This potential is what drives the current in an electrochemical cell.

One might liken it to a party where redox reactions are the guests and electrochemistry is the house where the party happens. Here, the Nernst Equation is like the rulebook that determines the potential energy snapshot of the party at any given moment, given certain guests (chemical species) have already decided to show up (concentration levels).

Standard Cell Potential

Think of standard cell potential (\(E^{\circ}\text{)\) as the default electrical push when conditions are perfect—1 molar concentration at 25°C and 1 atmosphere of pressure for all reactants and products. It’s a handy reference point. In our problem, the standard cell potential of \(+0.36 \mathrm{~V}\) is the starting line for the reaction when equal concentrations of oxidized and reduced forms are present.

However, life isn’t always standard. That's where the Nernst Equation comes into play, allowing us to adjust our expectations for cell potential when concentrations shift from that perfect 1 molar standard. It helps answer questions like, 'What happens to the voltage when we mess with the concentrations?'—a bit like tuning a guitar to keep the music flowing when the temperature or humidity changes.

Chemical Species Concentration

Concentration detailing how crowded a party is with different chemical species often dictates the direction and extent a chemical reaction will proceed. In the equation from the exercise, the concentrations of \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}\) and \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}\) directly influence the potential of the redox reaction. The Nernst Equation lets us quantify this influence.

It's akin to adjusting the volume of music in a room to suit your mood. Just like how you'd slide a volume fader up and down, you can modify reaction conditions, like concentration, to 'tune' the cell’s potential. The result from the Nernst Equation gives us a new potential that reflects these changes—information that's crucial for scientists and engineers when designing batteries and other electrochemical devices.