Respuesta :
Answer: The correct answer is true.
Explanation:
The given cell is:
[tex]Mn(s)/Mn^{2+}(1.04M)||Cu^{2+}(2.32\times 10^{-4}M)/Cu(s)[/tex]
Half reactions for the given cell follows:
Oxidation half reaction: [tex]Mn(s)\rightarrow Mn^{2+}(1.04M)+2e^-;E^o_{Mn^{2+}/Mn}=-1.18V[/tex]
Reduction half reaction: [tex]Cu^{2+}(2.32\times 10^{-4}M)+2e^-\rightarrow Cu(s);E^o_{Cu^{2+}/Cu}=0.34V[/tex]
Net reaction: [tex]Mn(s)+Cu^{2+}(2.32\times 10^{-4}M)\rightarrow Mn^{2+}(1.04M)+Cu(s)[/tex]
Oxidation reaction occurs at anode and reduction reaction occurs at 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}=0.34-(-1.18)=1.52V[/tex]
- To calculate the EMF of the cell, we use the Nernst equation, which is:
[tex]E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Mn^{2+}]}{[Cu^{2+}]}[/tex]
where,
[tex]E_{cell}[/tex] = electrode potential of the cell = ?V
[tex]E^o_{cell}[/tex] = standard electrode potential of the cell = +1.52 V
n = number of electrons exchanged = 2
[tex][Cu^{2+}]=2.32\times 10^{-4}M[/tex]
[tex][Mn^{2+}]=1.04M[/tex]
Putting values in above equation, we get:
[tex]E_{cell}=1.52-\frac{0.059}{2}\times \log(\frac{1.04}{2.32\times 10^{-4}})\\\\E_{cell}=1.412V[/tex]
As, the EMF of the cell is coming out to be positive. Thus, the cell reaction is spontaneous for the given concentration.
Hence, the correct answer is true.
