Question

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

Answer 1

E cell = +1.95 V

Step-by-step 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