Question

Use the following \(E^{\circ}\) for the electrode potentials, calculate \(\Delta G^{\circ}\) in \(\mathrm{kJ}\) for the indicated reaction : $$ \begin{aligned} 5 \mathrm{Ce}^{4+}(a q)+\mathrm{Mn}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l) & 5 \mathrm{Ce}^{3+}(a q) ; \quad+\mathrm{MnO}_{4}^{-}(a q)+8 \mathrm{H}^{+}(a q) \\ \mathrm{MnO}_{4}^{-}(a q)+8 \mathrm{H}^{+}(a q)+5 e^{-} & \mathrm{Mn}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l) ; \quad E^{\circ}=+1.51 \mathrm{~V} \end{aligned} $$ \(\mathrm{Ce}^{4+}(\alpha q)+\mathrm{e}^{-} \longrightarrow \mathrm{Ce}^{3+}(a q) \quad E^{\circ}=+1.61 \mathrm{~V}\) (a) \(-9.65\) (b) \(-24.3\) (c) \(-48.25\) (d) \(-35.2\)

Short answer

\(-48.25\, \mathrm{kJ}\) is the correct value for \(\Delta G^{\circ}\).

Step-by-step solution

Identify the Full Reaction

Combine the given half-reactions to understand the complete reaction for which \(\Delta G^{\circ}\) is to be calculated. The full reaction is obtained by adding the two half-reactions together, ensuring that the electrons cancel out.

Balance Electrons in Half-Reactions

Balance the electrons in the two half-reactions by multiplying each half-reaction by a suitable integer so that the number of electrons lost in the oxidation half is equal to the number of electrons gained in the reduction half.

Write the Net Equation

After balancing the electrons, write down the net equation of the reaction by adding the two balanced half-reactions, and ensure all species are balanced.

Calculate the Standard Electrode Potential

Use the formula \(E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}\) to find the potential of the cell from the given standard electrode potentials.

Calculate \(\Delta G^{\circ}\)

Finally, use the equation \(\Delta G^{\circ} = -nFE^{\circ}_{\text{cell}}\) to calculate the change in free energy, where \(n\) is the number of moles of electrons transferred, \(F\) is Faraday's constant \(96,485 \, C/mol\), and \(E^{\circ}_{\text{cell}}\) is the standard cell potential.

Key concepts

Standard Electrode Potential

Understanding standard electrode potential is essential when delving into electrochemistry. It's a measure of the tendency of a chemical species to lose or gain electrons and become reduced or oxidized. Think of it as an individual's savings account balance; it tells us how likely a person is to spend or save based on the balance.
Every half-cell in an electrochemical cell has its own standard electrode potential, denoted as \(E^\circ\), which is measured under standard conditions: a 1 Molar solution at 298 K temperature and 1 atmosphere of pressure. When considering the reaction given in the exercise, we look at the potential of cerium and manganese half-cells to calculate the overall cell potential.
This potential is crucial because it allows us to predict the direction of the electron flow in a cell. A higher potential encourages reduction (gaining electrons), while a lower potential means oxidation (losing electrons). So, by comparing the standard electrode potentials of two half-reactions, we can determine which one will act as the cathode (site of reduction) and which one as the anode (site of oxidation).

Gibbs Free Energy

Gibbs Free Energy, represented as \(G\), is a bit like a forecast that predicts whether a reaction will occur spontaneously. It’s an indicator of the maximum amount of work that can be performed by a thermodynamic system at a constant temperature and pressure.
The formula \(\Delta G^\circ = -nFE^\circ_{\text{cell}}\) joins the worlds of thermodynamics and electrochemistry. Here, \(\Delta G^\circ\) is the change in Gibbs Free Energy under standard conditions, \(n\) is the number of moles of electrons exchanged, \(F\) represents Faraday's constant, and \(E^\circ_{\text{cell}}\) is the standard cell potential calculated earlier.
Predicting the spontaneity of a reaction is like predicting whether a ball will roll downhill without any pushing - a negative \(\Delta G^\circ\) means the process will happen on its own, indicating a spontaneous reaction. In the given exercise, calculating \(\Delta G^\circ\) helps us understand the feasibility of the redox reaction described.

Redox Reactions

Redox reactions form the bedrock of electrochemistry. They involve the transfer of electrons between substances. Think of it as a dance between partners - one gives electrons (oxidation), while the other takes them (reduction).
Understanding the half-reactions is a key step. Each half-reaction represents either the loss (oxidation) or gain (reduction) of electrons. The balancing act is crucial here because the electrons lost in one half-reaction must be gained in another, ensuring that charge is conserved.
In the provided exercise, balancing the electrons allows us to stitch the two half-reactions into a full picture that aligns with the principle of electron conservation. By meticulously following the steps, we ensure that each electron that enters the dance leaves with a partner, maintaining the balance necessary for the reaction to proceed.

Nernst Equation

The Nernst Equation is like the GPS for navigating non-standard conditions in electrochemistry, helping us to adjust the standard electrode potential based on the concentration of reactants and products.
Although not directly applied in this particular exercise, understanding the Nernst Equation is useful for handling real-world scenarios where conditions aren't at the standard state. The equation relates the measurable cell potential to the standard electrode potential (\(E^\circ\)), temperature (\(T\)), number of moles of electrons transferred in the reaction (\(n\)), the charge on an electron (\(F\)), and the reaction quotient (\(Q\)).
It gives us insight into how potential changes with concentration, and it is absolutely critical for JEE aspirants to master its use to solve problems under varied conditions. This is a cornerstone concept in JEE Physical Chemistry and many times, the linchpin of understanding complex electrochemical systems.

JEE Physical Chemistry

Preparing for the Joint Entrance Examination (JEE) requires a deep dive into Physical Chemistry, which is a fusion of chemistry with physics principles. The topics of standard electrode potential, Gibbs Free Energy, redox reactions, and the Nernst Equation are central themes within this field.
Performing well in JEE Physical Chemistry demands not just memorizing equations and concepts, but also truly understanding how they interconnect and apply to solving real-world problems. The exercise provided showcases these principles in action, demonstrating how they're not merely theoretical concepts but tools that enable us to dissect and understand the flow of energy and matter at the molecular level.
Remember, consistency in practice and a clear grasp of fundamentals are your allies in tackling JEE Physical Chemistry effectively. By interpreting data, manipulating equations, and applying these principles, your problem-solving acumen will grow - making you well-prepared for the JEE.