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To calculate the standard Gibbs free energy change (ΔG°) for the given reaction, we need to use the concept of standard reduction potentials and the Gibbs free energy equation. Here’s the step-by-step solution:
1. Identify the half-reactions and their corresponding standard reduction potentials:
- Reduction half-reaction for [tex]\( \text{MnO}_4^- \)[/tex]:
[tex]\[ \text{MnO}_4^- + 8H^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4H_2O \][/tex]
with [tex]\( E^\circ = 1.51 \text{ V} \)[/tex]
- Oxidation half-reaction for [tex]\( \text{Br}^- \)[/tex]:
[tex]\[ 2\text{Br}^- \rightarrow \text{Br}_2 + 2e^- \][/tex]
The standard reduction potential for [tex]\(\text{Br}_2 + 2e^- \rightarrow 2\text{Br}^- \)[/tex] is [tex]\(1.07 \text{ V}\)[/tex], but for oxidation, it’s flipped to:
[tex]\[ E^\circ = -1.07 \text{ V} \][/tex]
2. Determine the overall cell potential (E°cell):
[tex]\[ E^\circ_\text{cell} = E^\circ_{\text{reduction}} - E^\circ_{\text{oxidation}} \][/tex]
which gives:
[tex]\[ E^\circ_\text{cell} = 1.51 \text{ V} - 1.07 \text{ V} = 0.44 \text{ V} \][/tex]
3. Calculate the Gibbs free energy change (ΔG°):
We use the formula:
[tex]\[ \Delta G^\circ = -nFE^\circ_\text{cell} \][/tex]
Where:
- [tex]\( n \)[/tex] = number of moles of electrons transferred in the balanced equation.
- [tex]\( F \)[/tex] = Faraday's constant = [tex]\( 96485 \text{ C/mol} \)[/tex].
From the balanced equation, we note that 10 [tex]\( \text{Br}^- \)[/tex] lose 10 electrons (since [tex]\(2 \text{MnO}_4^- \)[/tex] corresponds to 10 electrons), so [tex]\( n = 10 \)[/tex].
4. Substitute and calculate ΔG° in Joules:
[tex]\[ \Delta G^\circ = - (10 \text{ mol}) (96485 \text{ C/mol}) (0.44 \text{ V}) = -424533.99999999994 \text{ J} \][/tex]
5. Convert ΔG° into kJ (since 1 kJ = 1000 J):
[tex]\[ \Delta G^\circ = -424533.99999999994 \text{ J} / 1000 = -424.534 \text{ kJ} \][/tex]
6. Round the ΔG° to 3 significant digits:
[tex]\[ \Delta G^\circ \approx -424.534 \text{ kJ} \][/tex]
Thus, the standard Gibbs free energy change for the given reaction, rounded to 3 significant digits, is:
[tex]\[ \boxed{-424.534 \text{ kJ} \times 10 } \][/tex]
1. Identify the half-reactions and their corresponding standard reduction potentials:
- Reduction half-reaction for [tex]\( \text{MnO}_4^- \)[/tex]:
[tex]\[ \text{MnO}_4^- + 8H^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4H_2O \][/tex]
with [tex]\( E^\circ = 1.51 \text{ V} \)[/tex]
- Oxidation half-reaction for [tex]\( \text{Br}^- \)[/tex]:
[tex]\[ 2\text{Br}^- \rightarrow \text{Br}_2 + 2e^- \][/tex]
The standard reduction potential for [tex]\(\text{Br}_2 + 2e^- \rightarrow 2\text{Br}^- \)[/tex] is [tex]\(1.07 \text{ V}\)[/tex], but for oxidation, it’s flipped to:
[tex]\[ E^\circ = -1.07 \text{ V} \][/tex]
2. Determine the overall cell potential (E°cell):
[tex]\[ E^\circ_\text{cell} = E^\circ_{\text{reduction}} - E^\circ_{\text{oxidation}} \][/tex]
which gives:
[tex]\[ E^\circ_\text{cell} = 1.51 \text{ V} - 1.07 \text{ V} = 0.44 \text{ V} \][/tex]
3. Calculate the Gibbs free energy change (ΔG°):
We use the formula:
[tex]\[ \Delta G^\circ = -nFE^\circ_\text{cell} \][/tex]
Where:
- [tex]\( n \)[/tex] = number of moles of electrons transferred in the balanced equation.
- [tex]\( F \)[/tex] = Faraday's constant = [tex]\( 96485 \text{ C/mol} \)[/tex].
From the balanced equation, we note that 10 [tex]\( \text{Br}^- \)[/tex] lose 10 electrons (since [tex]\(2 \text{MnO}_4^- \)[/tex] corresponds to 10 electrons), so [tex]\( n = 10 \)[/tex].
4. Substitute and calculate ΔG° in Joules:
[tex]\[ \Delta G^\circ = - (10 \text{ mol}) (96485 \text{ C/mol}) (0.44 \text{ V}) = -424533.99999999994 \text{ J} \][/tex]
5. Convert ΔG° into kJ (since 1 kJ = 1000 J):
[tex]\[ \Delta G^\circ = -424533.99999999994 \text{ J} / 1000 = -424.534 \text{ kJ} \][/tex]
6. Round the ΔG° to 3 significant digits:
[tex]\[ \Delta G^\circ \approx -424.534 \text{ kJ} \][/tex]
Thus, the standard Gibbs free energy change for the given reaction, rounded to 3 significant digits, is:
[tex]\[ \boxed{-424.534 \text{ kJ} \times 10 } \][/tex]
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