Question

calculate the gibbs free energy for the following cell. graphite | h₂o, h⁺, o₂, cd²⁺ | cd half-reaction e° (v) cd²⁺ + 2e⁻ → cd -0.40 o₂ + 4h⁺ + 4e⁻ → 2h₂o + 1.23 e°_cell = -1.63 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 the number of 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 ($n$)

First, balance the half-reactions. The reduction of $\text{Cd}^{2+}$: $\text{Cd}^{2+} + 2\text{e}^-
ightarrow \text{Cd}$ (2 electrons gained). The reduction of $\text{O}_2$: $\text{O}_2 + 4\text{H}^+ + 4\text{e}^-
ightarrow 2\text{H}_2\text{O}$ (4 electrons gained). To balance the electrons, multiply the $\text{Cd}$ half-reaction by 2: $2\text{Cd}^{2+} + 4\text{e}^-
ightarrow 2\text{Cd}$. Now, the number of electrons transferred ($n$) is 4 (since both half-reactions now involve 4 electrons).

Step3: Substitute the values into the formula

We have $n = 4$, $F = 96.485\ \text{kJ/(V·mol e}^-)$, and $E^\circ_{cell} = -1.63\ \text{V}$.
Substitute into $\Delta G^\circ = -nFE^\circ_{cell}$:
$\Delta G^\circ = - (4\ \text{mol e}^-) \times (96.485\ \text{kJ/(V·mol e}^-)) \times (-1.63\ \text{V})$

Step4: Calculate the value

First, multiply the constants: $4 \times 96.485 = 385.94$. Then multiply by $-1.63$: $385.94 \times (-1.63) = -629.0822$. Then apply the negative sign from the formula: $-(-629.0822) = 629.0822\ \text{kJ}$. Wait, no, let's recheck. Wait, the formula is $\Delta G^\circ = -nFE^\circ_{cell}$. So:
$\Delta G^\circ = - (4) \times (96.485) \times (-1.63)$
$= 4 \times 96.485 \times 1.63$
Calculate $96.485 \times 1.63 = 157.27055$. Then $4 \times 157.27055 = 629.0822\ \text{kJ}$. Wait, but let's check the sign again. Wait, $E^\circ_{cell}$ is the cell potential. Wait, actually, in a galvanic cell, $E^\circ_{cell}$ is positive, but here $E^\circ_{cell} = -1.63\ \text{V}$, which means the reaction is non-spontaneous as written (it's an electrolytic cell or the reverse reaction is spontaneous). But the formula still holds. Wait, maybe I messed up the half-reactions. Wait, the cell notation is Graphite | $\text{H}_2\text{O}$, $\text{H}^+$, $\text{O}_2$, $\text{Cd}^{2+}$ | $\text{Cd}$. So the anode and cathode: the left side (Graphite) is the anode (oxidation) or cathode? Wait, in cell notation, the left is anode (oxidation), right is cathode (reduction). Wait, maybe I got the half-reactions reversed. Wait, the $\text{Cd}^{2+} + 2\text{e}^-
ightarrow \text{Cd}$ is reduction (cathode, right side). The $\text{O}_2$ reaction: if Graphite is the anode, then oxidation occurs there. Wait, maybe the oxidation is $\text{Cd}
ightarrow \text{Cd}^{2+} + 2\text{e}^-$ (reverse of the given $\text{Cd}$ half-reaction, so $E^\circ_{ox} = +0.40\ \text{V}$), and reduction is $\text{O}_2 + 4\text{H}^+ + 4\text{e}^-
ightarrow 2\text{H}_2\text{O}$ ($E^\circ_{red} = +1.23\ \text{V}$). Then $E^\circ_{cell} = E^\circ_{red} - E^\circ_{ox} = 1.23 - 0.40 = 0.83\ \text{V}$? Wait, no, the given $E^\circ_{cell} = -1.63\ \text{V}$. So maybe the cell is set up with the $\text{O}_2$ reduction as the anode? No, the problem states $E^\circ_{cell} = -1.63\ \text{V}$, so we have to use that. Let's go back. The formula is $\Delta G^\circ = -nFE^\circ_{cell}$. We need to find $n$. Let's balance the two half-reactions as they are. The $\text{Cd}$ half-reaction: 2 electrons. The $\text{O}_2$ half-reaction: 4 electrons. To make the electrons equal, multiply the $\text{Cd}$ half-reaction by 2: $2\text{Cd}^{2+} + 4\text{e}^-
ightarrow 2\text{Cd}$ (reduction, 4 electrons gained). The $\text{O}_2$ half-reaction: $\text{O}_2 + 4\text{H}^+ + 4\text{e}^-
ightarrow 2\text{H}_2\text{O}$ (reduction, 4 electrons gained). Wait, but in a cell, one is oxidation and one is reduction. Oh! I see my mistake. In a cell, one half-reaction is oxidation (reverse of reduction) and the other is reduction. So if the cell is Graphite | $\text{H}_2\text{O}$, $\text{H}^+$, $\text{O}_2$, $\text{Cd}^{2+}$ | $\text{Cd}$, the left electrode (Graphite) is the anode (oxidation), and the right (Cd) is the cathode (reduction). So the oxidation half-reaction is the reverse of the $\text{O}_2$ reduction? No, wait, the species at the anode: $\text{H}_2\text{O}$, $\text{H}^+$, $\text{O}_2$ – maybe the oxidation is $\text{H}_2\text{O}
ightarrow \text{O}_2 + 4\text{H}^+ + 4\text{e}^-$ (reverse of the $\text{O}_2$ reduction, so $E^\circ_{ox} = -1.23\ \text{V}$), and the reduction is $\text{Cd}^{2+} + 2\text{e}^-
ightarrow \text{Cd}$ ($E^\circ_{red} = -0.40\ \text{V}$). Then $E^\circ_{cell} = E^\circ_{red} + E^\circ_{ox} = -0.40 + (-1.23) = -1.63\ \text{V}$, which matches the given $E^\circ_{cell}$. So the oxidation half-reaction is $\text{2H}_2\text{O}
ightarrow \text{O}_2 + 4\text{H}^+ + 4\text{e}^-$ (4 electrons lost), and the reduction is $\text{2Cd}^{2+} + 4\text{e}^-
ightarrow 2\text{Cd}$ (4 electrons gained). So $n = 4$ (since 4 electrons are transferred). Now, $E^\circ_{cell} = -1.63\ \text{V}$, $n = 4$, $F = 96485\ \text{C/mol e}^-$. So $\Delta G^\circ = -nFE^\circ_{cell} = - (4\ \text{mol e}^-) \times (96485\ \text{C/mol e}^-) \times (-1.63\ \text{V})$. Since $1\ \text{J} = 1\ \text{C·V}$, and $1\ \text{kJ} = 1000\ \text{J}$, we can convert:

First, calculate the charge: $4 \times 96485 = 385940\ \text{C}$. Then multiply by $-1.63\ \text{V}$: $385940 \times (-1.63) = -629082.2\ \text{J}$. Then apply the negative sign from the formula: $-(-629082.2) = 629082.2\ \text{J} = 629.0822\ \text{kJ}$. Wait, but that can't be right. Wait, no, the formula is $\Delta G^\circ = -nFE^\circ_{cell}$. So if $E^\circ_{cell}$ is negative, then $\Delta G^\circ$ will be positive (since two negative signs: $-nF \times (negative) = positive$). Let's check the units: $F$ is in C/mol e⁻, $E^\circ_{cell}$ is in V (J/C), so $nF E^\circ_{cell}$ is in (mol e⁻) × (C/mol e⁻) × (J/C) = J. So converting to kJ, divide by 1000. So:

$n = 4$, $F = 96485\ \text{C/mol e}^-$, $E^\circ_{cell} = -1.63\ \text{V}$

$\Delta G^\circ = - (4) \times (96485) \times (-1.63) \times 10^{-3}\ \text{kJ}$ (since 1 J = 10⁻³ kJ)

Calculate:

$4 \times 96485 = 385940$

$385940 \times 1.63 = 385940 \times 1 + 385940 \times 0.63 = 385940 + 243142.2 = 629082.2$

Then multiply by $10^{-3}$: $629082.2 \times 10^{-3} = 629.0822\ \text{kJ}$

So $\Delta G^\circ = +629\ \text{kJ}$ (rounded to a reasonable number, or maybe 629 kJ). Wait, but let's check the calculation again. Wait, the formula is $\Delta G^\circ = -nFE^\circ_{cell}$. So:

$n = 4$, $F = 96485\ \text{C/mol}$, $E^\circ_{cell} = -1.63\ \text{V}$

$\Delta G^\circ = - (4) \times (96485) \times (-1.63) = 4 \times 96485 \times 1.63$

Calculate $96485 \times 1.63$:

$96485 \times 1 = 96485$

$96485 \times 0.6 = 57891$

$96485 \times 0.03 = 2894.55$

Sum: $96485 + 57891 = 154376 + 2894.55 = 157270.55$

Then $4 \times 157270.55 = 629082.2\ \text{J} = 629.0822\ \text{kJ} \approx 629\ \text{kJ}$ (or more precisely, 629 kJ). Wait, but maybe I made a mistake in $n$. Let's re-examine the half-reactions. The given half-reactions:

Reduction of $\text{Cd}^{2+}$: $\text{Cd}^{2+} + 2\text{e}^-
ightarrow \text{Cd}$ ($E^\circ = -0.40\ \text{V}$)

Reduction of $\text{O}_2$: $\text{O}_2 + 4\text{H}^+ + 4\text{e}^-
ightarrow 2\text{H}_2\text{O}$ ($E^\circ = +1.23\ \text{V}$)

If the cell is Graphite | $\text{H}_2\text{O}$, $\text{H}^+$, $\text{O}_2$, $\text{Cd}^{2+}$ | $\text{Cd}$, then the anode is Graphite (where oxidation occurs) and cathode is Cd (where reduction occurs). So the oxidation half-reaction is the reverse of the $\text{O}_2$ reduction: $\text{2H}_2\text{O}
ightarrow \text{O}_2 + 4\text{H}^+ + 4\text{e}^-$ ($E^\circ_{ox} = -1.23\ \text{V}$)

The reduction half-reaction is $\text{Cd}^{2+} + 2\text{e}^-
ightarrow \text{Cd}$ ($E^\circ_{red} = -0.40\ \text{V}$)

To balance the electrons, multiply the reduction half-reaction by 2: $2\text{Cd}^{2+} + 4\text{e}^-
ightarrow 2\text{Cd}$ ($E^\circ_{red} = -0.40\ \text{V}$)

Now, the overall reaction is: $2\text{H}_2\text{O} + 2\text{Cd}^{2+}
ightarrow \text{O}_2 + 4\text{H}^+ + 2\text{Cd}$

The number of electrons transferred ($n$) is 4 (since 4 electrons are lost in oxidation and 4 are gained in reduction). So $n = 4$ is correct.

Then, $\Delta G^\circ = -nFE^\circ_{cell} = - (4) \times (96485\ \text{C/mol}) \times (-1.63\ \text{V})$

$= 4 \times 96485 \times 1.63\ \text{J}$

$= 4 \times 96485 \times 1.63 \times 10^{-3}\ \text{kJ}$

$= 4 \times 96.485 \times 1.63\ \text{kJ}$ (since $96485\ \text{C/mol} = 96.485\ \text{kJ/(V·mol)}$)

$4 \times 96.485 = 385.94$

$385.94 \times 1.63 = 385.94 \times 1 + 385.94 \times 0.6 + 385.94 \times 0.03$

$= 385.94 + 231.564 + 11.5782$

$= 385.94 + 231.564 = 617.504 + 11.5782 = 629.0822\ \text{kJ}$

So the value of $\Delta G^\circ$ is $+629\ \text{kJ}$ (or approximately 629 kJ).

Answer:

+629 (or more precisely, 629.08, but rounded to a reasonable value)