Respuesta :
Answer:
[tex]Mn(s)|Mn^{2+}(aq)||Fe^{3+}(aq)|Fe(s)[/tex]
Explanation:
The given half reactions are:
[tex]Mn^{2+}(aq)+2e^{-}\rightarrow Mn(s).....E^{0}=-1.18V[/tex]
[tex]Fe^{3+}(aq)+3e^{-}\rightarrow Fe(s).....E^{0}=-0.036V[/tex]
where E° is the standard reduction potential.
The electrode with a higher (more positive) reduction potential serves as the cathode.
Here, E°(Fe3+/Fe) > E°(Mn2+/Mn)
Hence Fe3+/Fe = cathode
Mn2+/Mn = anode
As per convention, the cell notation is represented such that the anode half cell is on the left and cathode on the right. Both a separated by a salt bridge represented by the symbol ('||')
Cell notation:
[tex]Mn(s)|Mn^{2+}(aq)||Fe^{3+}(aq)|Fe(s)[/tex]
