Question

Calculate the standard free-energy change for the following reaction at 25 C. Standard reduction potentials can be found here.

Answer 1

Answer: The standard free energy change for the reaction will be -1308.54 kJ.

Explanation:

We are given a net redox reaction:

$2Au^{3+}(aq.)+3Zn(s)\rightarrow 2Au(s)+3Zn^{2+}(aq.)$

Here, gold is getting reduced and zinc is getting oxidized. So, the half cell reactions for the above net reaction are:

At Cathode: $Au^{3+}(aq.)+3e^-\rightarrow Au(s)$  × 2   $E^o_{Au^{3+}/Au}=1.50V$

At Anode:   $Zn(s)\rightarrow Au^{2+}(aq.)+2e^-$  × 3    $E^o_{Zn/Zn^{2+}}=0.76V$

To calculate the $E^o_{cell}$, we use the equation:

$E^o_{cell}$ = Standard oxidation potential of the oxidation half reaction +  Standard reduction potential of the reduction half reaction

Putting values in above equation:

$E^o_{cell}=0.76+1.50=2.26V$

The total electrons change for the redox reaction are 6.

To calculate he standard free energy change, we use the equation:

$\Delta G^o=-nFE^o_{cell}$

where,

n = Number of electron change = 6

F = Faraday's constant = 96500 F

$E^o_{cell}=2.26V$

Putting values in above equation, we get:

$\Delta G^o=-6\times 96500\times 2.26\\\\\Delta G^o=-1308540J=-1308.54kJ$

Hence, the standard free energy change for the reaction will be -1308.54 kJ.

Answer 2

n = 6
Ecell = E Au - E Zn
= 1.498 - 0.76
= 0.738

to find the standard free energy, use the formula:
ΔG = -nFEcell 
so,
ΔG = -(6)(96,500)(0.738)
= -427.3 kJ