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 ($E^\circ_{cell}$) and moles of electrons transferred ($n$) is $\Delta G^\circ = -nFE^\circ_{cell}$, where $F$ is Faraday's constant ($F = 96485\ \text{C/mol e}^- = 96.485\ \text{kJ/(V·mol e}^-)$).

Step2: Determine the number of moles of electrons transferred ($n$)

For the anode (oxidation) reaction: $\text{Mn}
ightarrow \text{Mn}^{2+} + 2\text{e}^-$ (since the reduction of $\text{Mn}^{2+}$ has 2 electrons, oxidation is reverse, so 2 electrons lost).
For the cathode (reduction) reaction: $\text{Fe}^{3+} + \text{e}^-
ightarrow \text{Fe}^{2+}$ (1 electron gained per reaction). To balance electrons, multiply the cathode reaction by 2: $2\text{Fe}^{3+} + 2\text{e}^-
ightarrow 2\text{Fe}^{2+}$. Now, electrons lost (anode) = electrons gained (cathode) = 2. So $n = 2$.

Step3: Plug in the values into the formula

We have $E^\circ_{cell} = 1.95\ \text{V}$, $n = 2$, $F = 96.485\ \text{kJ/(V·mol e}^-)$.
$\Delta G^\circ = -nFE^\circ_{cell} = -2 \times 96.485\ \text{kJ/(V·mol e}^-) \times 1.95\ \text{V}$

Step4: Calculate the value

First, calculate $2 \times 96.485 \times 1.95$:
$2 \times 96.485 = 192.97$
$192.97 \times 1.95 = 192.97 \times (2 - 0.05) = 192.97 \times 2 - 192.97 \times 0.05 = 385.94 - 9.6485 = 376.2915$
Then, $\Delta G^\circ = -376.2915\ \text{kJ} \approx -376\ \text{kJ}$ (or more precisely, -376.3 kJ, but let's check the calculation again with more precision):

$96485\ \text{C/mol} = 96485\ \text{J/(V·mol)} = 96.485\ \text{kJ/(V·mol)}$

$n = 2$, $E^\circ_{cell} = 1.95\ \text{V}$

$\Delta G^\circ = -2 \times 96485\ \text{J/(V·mol)} \times 1.95\ \text{V} = -2 \times 96485 \times 1.95\ \text{J/mol}$

Calculate $96485 \times 1.95$:

$96485 \times 2 = 192970$

$96485 \times 0.05 = 4824.25$

$192970 - 4824.25 = 188145.75$

Then, $2 \times 188145.75 = 376291.5\ \text{J/mol} = 376.2915\ \text{kJ/mol}$

So $\Delta G^\circ = -376.2915\ \text{kJ} \approx -376\ \text{kJ}$ (or -376 kJ when rounded to three significant figures, since $E^\circ_{cell}$ has three significant figures, $n$ is exact, $F$ is known precisely, so the answer should have three significant figures: -376 kJ)

Answer:

-376 (or -376.3, depending on rounding; the precise calculation gives approximately -376 kJ)