Answer: The standard free energy change for the reaction will be -1308.54 kJ.
Explanation:
We are given a net redox reaction:
[tex]2Au^{3+}(aq.)+3Zn(s)\rightarrow 2Au(s)+3Zn^{2+}(aq.)[/tex]
Here, gold is getting reduced and zinc is getting oxidized. So, the half cell reactions for the above net reaction are:
At Cathode: [tex]Au^{3+}(aq.)+3e^-\rightarrow Au(s)[/tex] × 2 [tex]E^o_{Au^{3+}/Au}=1.50V[/tex]
At Anode: [tex]Zn(s)\rightarrow Au^{2+}(aq.)+2e^-[/tex] × 3 [tex]E^o_{Zn/Zn^{2+}}=0.76V[/tex]
To calculate the [tex]E^o_{cell}[/tex], we use the equation:
[tex]E^o_{cell}[/tex] = Standard oxidation potential of the oxidation half reaction + Standard reduction potential of the reduction half reaction
Putting values in above equation:
[tex]E^o_{cell}=0.76+1.50=2.26V[/tex]
The total electrons change for the redox reaction are 6.
To calculate he standard free energy change, we use the equation:
[tex]\Delta G^o=-nFE^o_{cell}[/tex]
where,
n = Number of electron change = 6
F = Faraday's constant = 96500 F
[tex]E^o_{cell}=2.26V[/tex]
Putting values in above equation, we get:
[tex]\Delta G^o=-6\times 96500\times 2.26\\\\\Delta G^o=-1308540J=-1308.54kJ[/tex]
Hence, the standard free energy change for the reaction will be -1308.54 kJ.
The standard free energy of the reaction is -1307.382 kJ.
We must approach the problem in steps;
Obtain the E°cell as follows;
E°Au - E°Zn = +1.498 V - (-0.76 V) =2.258 V
Using the formula;
ΔG = -nFE°cell
Substituting values;
ΔG = -(6 * 96500 * 2.258)
ΔG = -1307.382 kJ
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