Write the equations relating \(\Delta G^{\circ}\) and \(K\) to the standard emf of a cell. Define all the terms.

Short Answer

Expert verified
\(\Delta G^{\circ} = -nFE^{\circ}\) and \(\Delta G^{\circ} = -RT \ln K\). Therefore, \(E^{\circ} = \frac{RT}{nF} \ln K\)

Step by step solution

01

Define the terms

Firstly, let's define these terms: \n\n- \(\Delta G^{\circ}\) is the standard Gibbs free energy change. It is the maximum useful work that can be obtained from a system at constant temperature and pressure. It gives information about the equilibrium and spontaneity of a chemical reaction.\n\n- \(K\) is the equilibrium constant, which measures the direction of a chemical reaction at equilibrium. A large value of \(K\) implies that the reaction product concentration is greater than the reactant concentration at equilibrium.\n\n- The standard emf of a cell, denoted as \(E^{\circ}\), is a measure of the driving force of an electrochemical reaction. The greater the value of \(E^{\circ}\), the more the reaction tends to proceed to completion.
02

Write the equation relating \(\Delta G^{\circ}\) and \(E^{\circ}\)

The Gibbs free energy change and the standard emf are related by the equation: \[\Delta G^{\circ} = -nFE^{\circ}\] Where \(n\) is the number of moles of electrons transferred in the reaction and \(F\) is the Faraday constant, the amount of electric charge in one mole of electrons.
03

Write the equation relating \(\Delta G^{\circ}\) and \(K\)

The standard Gibbs free energy change and the equilibrium constant are related by the equation: \[\Delta G^{\circ} = -RT \ln K\] Where \(R\) is the universal gas constant and \(T\) is the temperature in Kelvin.
04

Combine the equations

Combining steps 2 and 3, we can write an indirect relation between \(E^{\circ}\) and \(K\): \[-nFE^{\circ} = -RT \ln K\] After rearranging we will get \[E^{\circ} = \frac{RT}{nF} \ln K\]

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