Question

How many electrons would be transferred in either a voltaic or electrolytic cell that uses the following half reactions

Fe3+ (aq) + 3 e− ------> Fe (s) Eo = −0.036 V
Mg2+(aq) + 2 e− -------> Mg (s) Eo = −2.37 V

Answer 1

Numbers of electrons transferred in the electrolytic or voltaic cell is 6 electrons.

Step-by-step explanation:

`Fe^(3+) (aq) + 3 e^-\rightarrow Fe (s) ,E^o = -0.036 V`

`Mg^(2+)(aq) + 2 e^- \rightarrow Mg (s),E^o = -2.37 V`

The substance having highest positive reduction
`E^o` potential will always get reduced and will undergo reduction reaction.

Reduction : cathode

`Fe^(3+) (aq) + 3 e^-\rightarrow Fe (s) ,E^o = -0.036 V`..[1]

Oxidation: anode

`Mg(s)\rightarrow Mg^(2+)(aq) + 2 e^-,E^o = 2.37 V`..[2]

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the
`E^o_(cell)` of the reaction, we use the equation:

`E^o_(cell)=E^o_(red,cathode)-E^o_(red,anode)`

`E^o_(cell)=-0.036V-(-2.37 V)=2.334 V`

The overall reaction will be:

2 × [1] + 3 × [2] :

`2Fe^(3+) (aq) + 3Mg(s)+6e^-\rightarrow 2Fe (s)+3Mg^(2+)(aq)+6e^-`

Electrons on both sides will get cancelled :

`2Fe^(3+) (aq) + 3Mg(s)\rightarrow 2Fe (s)+3Mg^(2+)(aq)`

Numbers of electrons transferred in the electrolytic or voltaic cell is 6 electrons.