Animated Solution for Chemistry - Electrochemistry: An oxidation-reduction reaction in which 3 electrons are transferred has a $\Delta G^{\circ }$ of $17.37{\text{ kJ mol}}^{-1}$ at $25^{\circ }\text{C}$. The value of $E_{\text{cell}}^{\circ }$ (in V) is …… $\times 10^{-2}$. ($1F=96,500{\text{ C mol}}^{-1}$)

The relationship between thermodynamics and electrochemistry is one of the most beautiful intersections in physical chemistry. It allows us to predict whether a chemical reaction can generate electricity, or if we need to pump electricity into the system to force the reaction to happen.

In this problem, we are given the standard Gibbs free energy change ($\Delta G^{\circ }$) for an oxidation-reduction reaction and asked to find the standard cell potential ($E_{\text{cell}}^{\circ }$). Let's dive into the mechanics of this calculation.

The bridge between the chemical energy of a reaction and the electrical potential it can produce is given by the master equation:

Here, $\Delta G^{\circ }$ is the standard Gibbs free energy change, $n$ is the number of moles of electrons transferred in the balanced redox equation, $F$ is the Faraday constant (the charge of one mole of electrons, approximately $96500{\text{ C mol}}^{-1}$), and $E_{\text{cell}}^{\circ }$ is the standard cell potential.

Before we rush into plugging numbers into our equation, we must inspect our units. This is where many students make a critical error.

The given standard Gibbs free energy change is:

However, the Faraday constant ($F$) is given in Coulombs per mole. Since $1\text{ Volt}\times 1\text{ Coulomb}=1\text{ Joule}$, the energy term in our equation must be in Joules, not kilojoules.

Let's convert $\Delta G^{\circ }$ to Joules:

Now that our units are consistent, we can substitute the known values into the master equation. We know that $n=3$ electrons are transferred.

To isolate $E_{\text{cell}}^{\circ }$, we divide the energy by the total charge transferred:

Calculating the denominator:

Now, perform the final division:

The question specifically asks for the value of $E_{\text{cell}}^{\circ }$ in the format of $\cdots \times 10^{-2}$.

We can rewrite $-0.06$ as:

Thus, the integer value required for the answer is $-6$.

Take a moment to look at the signs. We started with a positive $\Delta G^{\circ }$ ($+17.37{\text{ kJ mol}}^{-1}$), which tells us that the reaction is non-spontaneous under standard conditions.

Consequently, our calculated $E_{\text{cell}}^{\circ }$ is negative ($-0.06\text{ V}$). A negative cell potential confirms that the cell cannot act as a galvanic cell (it won't produce electricity). Instead, it must operate as an electrolytic cell, requiring an external voltage of at least $0.06\text{ V}$ to drive the reaction forward. The math perfectly mirrors the physical reality!