Respuesta :
Answer: The concentration of [tex]Au^{3+}[/tex] in the solution is [tex]1.87\times 10^{-14}M[/tex]
Explanation:
The given cell is:
[tex]Pt(s)|H_2(g.1atm)|H^+(aq.,1.0M)||Au^{3+}(aq,?M)|Au(s)[/tex]
Half reactions for the given cell follows:
Oxidation half reaction: [tex]H_2(g)\rightarrow 2H^{+}(1.0M)+2e^-;E^o_{H^+/H_2}=0V[/tex] ( × 3)
Reduction half reaction: [tex]Au^{3+}(?M)+3e^-\rightarrow Au(s);E^o_{Au^{3+}/Au}=1.50V[/tex] ( × 2)
Net reaction: [tex]3H_2(s)+2Au^{3+}(?M)\rightarrow 6H^{+}(1.0M)+2Au(s)[/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]
Putting values in above equation, we get:
[tex]E^o_{cell}=1.50-0=1.50V[/tex]
To calculate the concentration of ion for given EMF of the cell, we use the Nernst equation, which is:
[tex]E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[H^{+}]^6}{[Au^{3+}]^2}[/tex]
where,
[tex]E_{cell}[/tex] = electrode potential of the cell = 1.23 V
[tex]E^o_{cell}[/tex] = standard electrode potential of the cell = +1.50 V
n = number of electrons exchanged = 6
[tex][Au^{3+}]=?M[/tex]
[tex][H^{+}]=1.0M[/tex]
Putting values in above equation, we get:
[tex]1.23=1.50-\frac{0.059}{6}\times \log(\frac{(1.0)^6}{[Au^{3+}]^2})[/tex]
[tex][Au^{3+}]=1.87\times 10^{-14}M[/tex]
Hence, the concentration of [tex]Au^{3+}[/tex] in the solution is [tex]1.87\times 10^{-14}M[/tex]
