Question

Calculate the standard free-energy change for the reaction at 25 ∘ C. 25 ∘C. Refer to the list of standard reduction potentials. 2 Au 3 + (aq) + 3 Ni (s) − ⇀ ↽ − 2 Au (s) + 3 Ni 2 + (aq) 2Au3+(aq)+3Ni(s)↽−−⇀2Au(s)+3Ni2+(aq) Δ G ∘ = ΔG∘=

Answer 1

Answer: The $\Delta G^o$ for the given reaction is $-1.02\times 10^6J$

Explanation:

For the given chemical reaction:

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

Here, gold is getting reduced because it is gaining electrons and nickel is getting oxidized because it is loosing electrons.

Oxidation half reaction:  $Ni(s)\rightarrow Ni^{2+}(aq.)+2e^-$   ( × 3)

Reduction half reaction:  $Au^{3+}(aq.)+3e^-\rightarrow Au(s)$   ( × 2)

We know that:

$E^o_{(Au^{3+}/Au)}=1.50V\\E^o_{(Ni^{2+}/Ni)}=-0.23V$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

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

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

$E^o_{cell}=1.50-(-0.23)=1.73V$

Relationship between standard Gibbs free energy and standard electrode potential follows:

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

where,

n = number of electrons transferred = 6

F = Faraday's constant = 96500 C

$E^o_{cell}$ = standard cell potential = 1.73 V

Putting values in above equation, we get:

$\Delta G^o=-6\times 96500\times 1.73=-1001670J=-1.02\times 10^6J$

Hence, the $\Delta G^o$ for the given reaction is $-1.02\times 10^6J$