Question

Consider a galvanic cell based on the reaction Al^3+_(aq) + Mg_(s) rightarrow Al_(s) + Mg^2+ _(aq) The half-reactions are Al^3+ + 3 e^- rightarrow Al E degree = - 1.66 V Mg^2+ + 2 e^- rightarrow Mg E degree = - 2.37 V Give the balanced cell reaction and calculate E degree for the cell.

Answer 1

Answer: The standard cell potential of the cell is -0.71 V

Step-by-step explanation:

The half reactions follows:

Oxidation half reaction:
`Mg\rightarrow Mg^(2+)+2e^-;E^o_(Mg^(2+)/Mg)=-2.37V` ( × 3)

Reduction half reaction:
`Al^(3+)(aq.)+3e^-\rightarrow Al(s);E^o_(Al^(3+)/Al)=-1.66V` ( × 2)

The balanced cell reaction follows:

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

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

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

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

Putting values in above equation, we get:

`E^o_(cell)=-2.37-(-1.66)=-0.71V`

Hence, the standard cell potential of the cell is -0.71 V