Answer:
The concentration [Au⁺³] is 0.287 M
Explanation:
The Nernst equation is useful for finding the potential for reduction in electrodes under conditions other than standards. This is what happens in this case, since the concentrations are different than 1M and the gas pressure varies from the value 1 atm.
The Nernst equation is:
[tex]E=E^{o} -\frac{R*T}{n*F} *ln(Q)[/tex] Equation (A)
Where E refers to the electrode potential.
Eº = potential in standard conditions.
R = gas constant.
T = absolute temperature (in Kelvin degrees).
n = number of moles that have participation in the reaction.
F = Faraday constant (with a value of 96500 C / mol, approx.)
Q = reaction ratio
For the reaction aA + bB → cC + dD, Q adopts the expression:
[tex]Q =\frac{[C] ^ c. [D] ^ d}{[A] ^ a. [B] ^ b}[/tex]
Taking into account that R = 8.31 J. mol-1 K-1, T = 25 C = 298.15 K (with 0 C = 273.15 K):
[tex]E=E^{o} -\frac{0.0592}{n} *ln(Q)[/tex] Equation (B)
The reduction reaction (chemical reaction from which an atom, ion or molecule gives up electrons and increases its oxidation state) will be the one with the greatest reduction potential. The oxidation reaction (reaction in which a substance captures electrons) is one with a lower reduction potential. In this case:
Reduction reaction: Au⁺³ + 3e⁻ ⇒ Au
Oxidation reaction: Cu ⇒ Cu⁺² + 2e⁻
Writing the global reaction equaling the number of electrons:
2 Au⁺³ + 3 Cu ⇒ 2 Au + 3 Cu⁺²
Then E⁰=Ereduction - Eoxidation= 1.5 V - 0.34 V= 1.16 V
So, being:
Replacing in Equation (B):
[tex]1.13 V=1.16 V-\frac{0.0592}{6} *ln(\frac{1.20^{3} }{[Au^{+3}]^{2} })[/tex]
Solving, you get:
[Au⁺³]= 0.287 M
The concentration [Au⁺³] is 0.287 M
