Question

Fe2+(aq)|Fe(s) -0.44 V and Zn2+(aq)|Zn(s) -0.76 V and Cr3+(aq)|Cr(s) ‑0.74 V and Cu2+(aq)/Cu(s) +0.34 V. Calculate the standard free energy change: Fe2+(aq) + Cr(s) ---> Fe(s) + Cr3+(aq)?

Answer 1

I think you are right and here's why:

Explanation:

`sf(Cr^(3+)+3erightleftharpoonsCr" "E^@=-0.74" "V)`

`sf(Fe^(2+)+2erightleftharpoonsFe" "E^@=-0.44" "V)`

This means the 1st 1/2 cell will be driven right to left and the 2nd 1/2 cell left to right.

The cell reaction will be:

`sf(3Fe^(2+)+2Crrarr3Fe+2Cr^(3+))`

To find `sf(E_(cell)^@)` subtract the least +ve value from the most +ve:

`sf(E_(cell)^@=-0.44-(-0.77)=+0.3color(white)(x)V)`

The free energy change is given by:

`sf(DeltaG^@=-nFE_(cell)^@)`

Where n is the number of moles of electrons transferred which, in this case = 6.

`:.` `sf(DeltaG^@=-6xx9.65xx10^(4)xx0.3=-173.7color(white)(x)kJ)`

Free energy change is an extensive quantity so depends on the amount of reactants and products.

The equation in the question is not balanced. This value refers to the equation as I have written it.

The answer key gives +683.1 kJ. Does it specify the equation to which it refers?

The + sign means the reaction is non - spontaneous. This is not the case indicated by the -ve sign of `sf(DeltaG^@)`.