Question

Explain why cell potentials are not multiplied by the coefficients in the balanced redox equation. (Use the relationship between \(\Delta G\) and cell potential to do this.)

Short answer

In conclusion, cell potentials are not multiplied by the coefficients in the balanced redox equation because they are intensive properties, meaning they do not depend on the amounts of substances involved in the reaction. Additionally, the relationship between Gibbs Free Energy (ΔG) and cell potential (E) is given by ΔG = -nFE, which does not involve the stoichiometric coefficients. When the entire redox equation is multiplied by a factor k, the cell potential E remains unchanged, and the new ΔG is simply k times the original ΔG.

Step-by-step solution

Recall the relationship between ΔG and cell potential E

The relationship between Gibbs Free Energy (ΔG) and cell potential (E) is given by the following formula: \[ΔG = -nFE\] where: - ΔG is the Gibbs Free Energy (in joules or J), - n is the number of moles of electrons transferred in the redox reaction, - F is the Faraday's constant (96,485 C/mol), and - E is the cell potential (in volts or V).

Understand the meaning of cell potential E

The cell potential E is the difference in potential energy between the two half-reactions of a redox reaction. It determines the capacity of the redox reaction to do work. E is an intensive property, which means it does not depend on the amount or extent of the substances involved in the reaction. It is measured in volts (V) and is independent of the coefficients in the balanced redox equation.

Show how coefficients in the balanced redox equation affect ΔG

Now, let's analyze a generic balanced redox equation, given by: \(aA + bB \rightarrow cC + dD\) According to the stoichiometric coefficients (a, b, c, and d), the reaction can be written in terms of two half-reactions: Reduction: \(aA + ne^{-} \rightarrow bB\) Oxidation: \(cC \rightarrow dD + ne^{-}\) Now, let's calculate the Gibbs Free Energy change (ΔG) for the entire redox reaction. If we multiply the balanced redox equation by any factor k, the reaction would become: \(kaA + kbB \rightarrow kcC + kdD\) Notice that when we multiply the entire redox equation by k, it affects the number of moles of electrons transferred in the reaction as well. The new n would be kn. The ΔG for the new reaction is given by: \[ΔG' = -k \cdot n \cdot F \cdot E\] Since k is constant, and E remains the same (as it does not depend on the coefficients in the balanced redox equation), the new ΔG' is just k times the original ΔG: \[ΔG' = k \cdot ΔG\]

Conclude why cell potentials are not multiplied by coefficients in the balanced redox equation

As shown above, the cell potential E remains the same for the redox reaction despite the coefficients being multiplied by k. This is because E is an intensive property that is independent of the amounts of substances involved in the reaction. Also, the relationship between ΔG and E, given by ΔG = -nFE, is not affected by the stoichiometric coefficients in the balanced redox equation. Thus, it is clear that cell potentials are not multiplied by the coefficients in the balanced redox equation due to their independence from the amounts of substances and their direct relationship with ΔG.