Question

calculate the gibbs free energy for the following cell.
half-reaction | e° (v)
cl₂ + 2e⁻ → 2cl⁻ | +1.36
br₂ + 2e⁻ → 2br⁻ | + 1.09
δg° = ? kj
enter either a + or - sign and the magnitude in the answer.

Explanation:

Step1: Identify the anode and cathode reactions

In a galvanic cell, the reaction with the higher standard reduction potential (E°) is the reduction (cathode), and the one with the lower E° is oxidized (anode, reverse the reaction).

• Cathode (reduction): $\ce{Cl_{2} + 2e^{-} -> 2Cl^{-}}$ with $E^{\circ}_{cathode} = +1.36\ \text{V}$
• Anode (oxidation, reverse the $\ce{Br_{2} + 2e^{-} -> 2Br^{-}}$ reaction): $\ce{2Br^{-} -> Br_{2} + 2e^{-}}$ with $E^{\circ}_{anode} = - 1.09\ \text{V}$ (since we reverse the reduction reaction, the potential sign flips)

Step2: Calculate the cell potential ($E^{\circ}_{cell}$)

The formula for $E^{\circ}_{cell}$ is $E^{\circ}_{cell}=E^{\circ}_{cathode}-E^{\circ}_{anode}$ (or $E^{\circ}_{cathode}+|E^{\circ}_{anode}|$ when anode reaction is reversed).
$E^{\circ}_{cell}=1.36\ \text{V}-1.09\ \text{V}= 0.27\ \text{V}$

Step3: Recall the formula for Gibbs free energy change ($\Delta G^{\circ}$)

The formula is $\Delta G^{\circ}=-nFE^{\circ}_{cell}$, where:

• $n$ is the number of moles of electrons transferred (here, $n = 2$ as seen from the half - reactions, 2 moles of electrons are transferred)
• $F$ is Faraday's constant, $F = 96485\ \text{C/mol}$ (coulombs per mole of electrons)
• $E^{\circ}_{cell}$ is the standard cell potential in volts.

Step4: Substitute the values into the formula

First, calculate the value in joules and then convert to kilojoules.
$\Delta G^{\circ}=-nFE^{\circ}_{cell}$
Substitute $n = 2$, $F = 96485\ \text{C/mol}$, and $E^{\circ}_{cell}=0.27\ \text{V}$ (note that $1\ \text{V}=1\ \text{J/C}$)
$\Delta G^{\circ}=-2\times96485\ \text{C/mol}\times0.27\ \text{V}$
$\Delta G^{\circ}=-2\times96485\times0.27\ \text{J/mol}$
Calculate $2\times96485\times0.27 = 2\times26050.95=52101.9\ \text{J/mol}$
So, $\Delta G^{\circ}=- 52101.9\ \text{J/mol}\approx - 52\ \text{kJ/mol}$ (when rounded to two significant figures, or more precisely - 52.1 kJ, but let's check the calculation again for more precision)

Wait, let's do the calculation more accurately:
$96485\times2 = 192970$
$192970\times0.27=192970\times\frac{27}{100}=192970\times0.2 + 192970\times0.07=38594+13507.9 = 52101.9\ \text{J}$
Convert to kJ: $\frac{52101.9\ \text{J}}{1000}=52.1019\ \text{kJ}$
So $\Delta G^{\circ}=- 52.1\ \text{kJ}$ (or approximately - 52 kJ, but let's see if we need to be more precise)

Answer:

• 52 (or more precisely - 52.1, but - 52 is a reasonable answer considering significant figures. The half - reaction potentials have two decimal places, and the calculation gives us around - 52 kJ when rounded appropriately)