Answer:
E cell = +1.95 V
Explanation:
At Anode : Oxidation reaction takes place
At Cathode : Reduction reaction takes place
The reaction with lower value of reduction potential will undergo Oxidation
[tex]Mn^{2+}(aq)+2e^{-}\rightarrow Mn(s)[/tex] E = -1.18 V
This equation undergo oxidation reaction and become:
Anode(Oxidation-Half) :
[tex]Mn(s)\rightarrow Mn^{2+}(aq)+2e^{-}[/tex] E = +1.18 V
Cathode(Reduction-Half) :
[tex]Fe^{3+}(aq)+e^{-}\rightarrow Fe^{2+}[/tex] E =+0.77 V
To balance the reaction multiply reduction-Half with 2.We get :
[tex]Fe^{3+} +2Mn(s)+\rightarrow Fe^{2+} + 2Mn^{2+}[/tex]
Note that E is intensive property , do not multiply E of oxidation-half with 2
Ecell = 0.77 -(-1.18)
E = +1.95 V
