Question

An electrochemical cell is constructed using electrodes with the following half-cell reactions.

Answer 1

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

$Mn^{2+}(aq)+2e^{-}\rightarrow Mn(s)$   E = -1.18 V

This equation undergo oxidation reaction and become:

Anode(Oxidation-Half) :

$Mn(s)\rightarrow Mn^{2+}(aq)+2e^{-}$      E = +1.18 V

Cathode(Reduction-Half) :

$Fe^{3+}(aq)+e^{-}\rightarrow Fe^{2+}$    E =+0.77 V

To balance the reaction multiply reduction-Half with 2.We get :

$Fe^{3+} +2Mn(s)+\rightarrow Fe^{2+} + 2Mn^{2+}$

Note that E is intensive property , do not multiply E  of oxidation-half with 2

Ecell =  0.77 -(-1.18)

E = +1.95 V