Respuesta :
Answer: The [tex]\Delta G^o[/tex] for the given reaction is [tex]-1.02\times 10^6J[/tex]
Explanation:
For the given chemical reaction:
[tex]2Au^{3+}(aq.)+3Ni(s)\rightarrow 3Ni^{2+}(aq.)+2Au(s)[/tex]
Here, gold is getting reduced because it is gaining electrons and nickel is getting oxidized because it is loosing electrons.
Oxidation half reaction: [tex]Ni(s)\rightarrow Ni^{2+}(aq.)+2e^-[/tex] ( × 3)
Reduction half reaction: [tex]Au^{3+}(aq.)+3e^-\rightarrow Au(s)[/tex] ( × 2)
We know that:
[tex]E^o_{(Au^{3+}/Au)}=1.50V\\E^o_{(Ni^{2+}/Ni)}=-0.23V[/tex]
Substance getting oxidized always act as anode and the one getting reduced always act as cathode.
To calculate the [tex]E^o_{cell}[/tex] of the reaction, we use the equation:
[tex]E^o_{cell}=E^o_{cathode}-E^o_{anode}[/tex]
[tex]E^o_{cell}=1.50-(-0.23)=1.73V[/tex]
Relationship between standard Gibbs free energy and standard electrode potential follows:
[tex]\Delta G^o=-nFE^o_{cell}[/tex]
where,
n = number of electrons transferred = 6
F = Faraday's constant = 96500 C
[tex]E^o_{cell}[/tex] = standard cell potential = 1.73 V
Putting values in above equation, we get:
[tex]\Delta G^o=-6\times 96500\times 1.73=-1001670J=-1.02\times 10^6J[/tex]
Hence, the [tex]\Delta G^o[/tex] for the given reaction is [tex]-1.02\times 10^6J[/tex]
The standard free energy is - 1002 kJ.
Recall the formula;
ΔG∘ = -nFE∘cell
- ΔG∘ = change in free energy
- n = number of moles of electrons transferred
- F = Faraday constant
- E∘cell = standard cell potential
E∘cell = 1.50 V - (-0.23 V) = 1.73 V
Applying the formula above;
ΔG∘= -(6 × 96500 × 1.73 V)
ΔG∘= - 1002 kJ
Learn more about redox reaction: https://brainly.com/question/13978139
