Respuesta :
Answer: The cell potential of the cell is 0.31 V
Explanation:
We know that:
[tex]E^o_{(Br_2/2Br^-)}=1.07V\\E^o_{(MnO_4^-/Mn^{2+})}=1.51V[/tex]
The substance having highest positive [tex]E^o[/tex] potential will always get reduced and will undergo reduction reaction. Here, [tex]MnO_4^-[/tex] will undergo reduction reaction will get reduced. And, bromine will get oxidized.
Oxidation half reaction: [tex]2Br^-(aq.)+2e^-\rightarrow Br_2(l);E^o_{(Br_2^/2Br^-)}=1.07V[/tex] ( × 5)
Reduction half reaction: [tex]MnO_4^-(aq.)+8H^+(aq.)+5e^-\rightarrow Mn^{2+}(aq.)+4H_2O(l);E^o_{(MnO_4^-/Mn^{2+})}=1.51V[/tex] ( × 2)
The net reaction follows:
[tex]2MnO_4^-(aq.)+10Br^-(aq.)+16H^+(aq.)\rightarrow 2Mn^{2+}(aq.)+5Br_2(l)+8H_2O(l)[/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}=1.51-(1.07)=0.44V[/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+}]^2}{[MnO_4^{-}]^2\times [Br^-]^{10}\times [H^+]^{16}}[/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 = 10
[tex][H^{+}]=1M[/tex]
[tex][Mn^{2+}]=0.15M[/tex]
[tex][MnO_4^{-}]=0.01M[/tex]
[tex][Br^{-}]=0.01M[/tex]
Putting values in above equation, we get:
[tex]E_{cell}=0.44-\frac{0.059}{10}\times \log(\frac{(0.15)^2}{(0.01)^2\times (0.01)^{10}\times (1)^{16}})\\\\E_{cell}=0.31V[/tex]
Hence, the cell potential of the cell is 0.31 V
