Respuesta :
Answer:
For a: The reactions are written below.
For b: The standard cell potential of the given cell is +0.44 V.
For c: The cell potential for the given values is 0.408 V
For d: The cell potential for the given values is 0.4724 V.
Explanation:
The given cell is:
[tex]Mn(s)/Mn^{2+}||Cr^{3+}/Cr(s)[/tex]
- For a:
Half reactions for the given cell follows:
Oxidation half reaction: [tex]Mn(s)\rightarrow Mn^{2+}(aq.)+2e^-;E^o_{Mn^{2+}/Mn}=-1.18V[/tex] ( × 3)
Reduction half reaction: [tex]Cr^{3+}(aq.)+3e^-\rightarrow Cr(s);E^o_{Cr^{3+}/Cr}=-0.74V[/tex] ( × 2)
Net reaction: [tex]Mn(s)+Cr^{3+}(aq.)\rightarrow Mn^{2+}(aq.)+Cr(s)[/tex]
- For b:
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}=-0.74-(-1.18)=0.44V[/tex]
Hence, the standard cell potential of the given cell is 0.44 V
- For c:
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+}]}{[Cr^{3+}]}[/tex]
where,
[tex]E_{cell}[/tex] = electrode potential of the cell = ?V
[tex]E^o_{cell}[/tex] = standard electrode potential of the cell = +0.44 V
n = number of electrons exchanged = 6
[tex][Cr^{3+}]=1.0\times 10^{-4}M[/tex]
[tex][Mn^{2+}]=0.20M[/tex]
Putting values in above equation, we get:
[tex]E_{cell}=0.44-\frac{0.059}{6}\times \log(\frac{0.20}{1.0\times 10^{-4}})\\\\E_{cell}=0.408V[/tex]
Hence, the EMF of the cell for the given concentration is 0.408 V.
- For d:
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+}]}{[Cr^{3+}]}[/tex]
where,
[tex]E_{cell}[/tex] = electrode potential of the cell = ?V
[tex]E^o_{cell}[/tex] = standard electrode potential of the cell = +0.44 V
n = number of electrons exchanged = 6
[tex][Cr^{3+}]=0.20M[/tex]
[tex][Mn^{2+}]=1.0\times 10^{-4}M[/tex]
Putting values in above equation, we get:
[tex]E_{cell}=0.44-\frac{0.059}{6}\times \log(\frac{1.0\times 10^{-4}}{0.20})\\\\E_{cell}=0.4724V[/tex]
Hence, the EMF of the cell for the given concentration is 0.4724 V.
