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 Oxidation and Reduction

In a galvanic cell, the half - reaction with higher \(E^{\circ}\) is the reduction reaction, and the one with lower \(E^{\circ}\) is the oxidation reaction (reverse of the given half - reaction).
Reduction: \(Cl_{2}+2e^{-}
ightarrow2Cl^{-}\), \(E^{\circ}_{red}= + 1.36\ V\)
Oxidation: \(2Br^{-}
ightarrow Br_{2}+2e^{-}\), \(E^{\circ}_{ox}=- 1.09\ V\) (reverse of \(Br_{2}+2e^{-}
ightarrow2Br^{-}\))

Step2: Calculate Cell Potential (\(E^{\circ}_{cell}\))

The formula for \(E^{\circ}_{cell}\) is \(E^{\circ}_{cell}=E^{\circ}_{red}(cathode)-E^{\circ}_{red}(anode)\) (or \(E^{\circ}_{cell}=E^{\circ}_{red}+E^{\circ}_{ox}\)).
\(E^{\circ}_{cell}=E^{\circ}_{red}(Cl_{2}/Cl^{-})-E^{\circ}_{red}(Br_{2}/Br^{-})\)
\(E^{\circ}_{cell}=1.36\ V - 1.09\ V=0.27\ V\)

Step3: Determine the number of moles of electrons (\(n\))

From the half - reactions, the number of moles of electrons transferred (\(n\)) is 2 (since in both half - reactions, 2 moles of electrons are involved).

Step4: Calculate Gibbs Free Energy (\(\Delta G^{\circ}\))

The formula for \(\Delta G^{\circ}\) is \(\Delta G^{\circ}=-nFE^{\circ}_{cell}\), where \(F = 96485\ C/mol\) (Faraday's constant), \(n\) is the number of moles of electrons, and \(E^{\circ}_{cell}\) is the cell potential.
First, substitute the values:
\(n = 2\), \(F=96485\ C/mol\), \(E^{\circ}_{cell}=0.27\ V\) (Note: \(1\ V = 1\ J/C\))
\(\Delta G^{\circ}=-2\times96485\ C/mol\times0.27\ V\)
\(\Delta G^{\circ}=-2\times96485\times0.27\ J/mol\) (since \(V\times C=J\))
Calculate the value:
\(2\times96485\times0.27 = 2\times26050.95=52101.9\ J/mol = 52.1019\ kJ/mol\)
So, \(\Delta G^{\circ}=- 52.1\ kJ\) (rounded to a reasonable number of significant figures)

Answer:

\(- 52\) (or more precisely \(-52.1\))