Question

calculate the gibbs free energy for the following cell. mn | mn²⁺ || fe³⁺, fe²⁺ | pt half-reaction e° (v) mn²⁺ + 2e⁻ → mn -1.18 fe³⁺ + e⁻ → fe²⁺ +0.77 e°_cell = +1.95 v δg° = ? kj enter either a + or - sign and the magnitude in the answer.

Explanation:

Step1: Recall the formula for Gibbs Free Energy

The formula relating Gibbs Free Energy change ($\Delta G^\circ$) to cell potential ($E^\circ_{cell}$) is $\Delta G^\circ = -nFE^\circ_{cell}$, where $n$ is the number of moles of electrons transferred, $F$ is Faraday's constant ($F = 96485\ C/mol$ or $96485\ J/(V\cdot mol)$), and $E^\circ_{cell}$ is the standard cell potential.

Step2: Determine the value of $n$

First, we need to balance the redox reaction. The oxidation half - reaction (Mn is oxidized because its oxidation state increases from 0 in Mn to +2 in $Mn^{2+}$): $Mn
ightarrow Mn^{2+}+2e^-$ (loses 2 electrons). The reduction half - reaction: $Fe^{3+}+e^-
ightarrow Fe^{2+}$ (gains 1 electron). To balance the number of electrons, we multiply the reduction half - reaction by 2: $2Fe^{3+}+2e^-
ightarrow 2Fe^{2+}$. Now, when we add the oxidation and the balanced reduction half - reactions: $Mn + 2Fe^{3+}
ightarrow Mn^{2+}+2Fe^{2+}$. The number of moles of electrons transferred, $n = 2$.

Step3: Substitute the values into the formula

We know that $E^\circ_{cell}=1.95\ V$, $n = 2$, and $F = 96485\ J/(V\cdot mol)$.
$\Delta G^\circ=-nFE^\circ_{cell}$
$\Delta G^\circ=-2\times96485\ J/(V\cdot mol)\times1.95\ V$
First, calculate $2\times96485\times1.95$:
$2\times96485 = 192970$
$192970\times1.95=192970\times(2 - 0.05)=192970\times2-192970\times0.05 = 385940-9648.5 = 376291.5$
So, $\Delta G^\circ=- 376291.5\ J/mol$
Convert from joules to kilojoules by dividing by 1000: $\Delta G^\circ=-376.2915\ kJ/mol\approx - 376\ kJ/mol$ (or more precisely, we can keep more decimal places during calculation)

Answer:

• 376 (or more precisely, - 376.3 if we use more precise calculation. The value can also be calculated as $-2\times96485\times1.95\times10^{- 3}\ kJ/mol=-2\times96.485\times1.95\ kJ/mol=-192.97\times1.95\ kJ/mol=-376.2915\ kJ/mol\approx - 376\ kJ/mol$)