Question

What is the Std. Gibbs free energy change and max. work obainable for reacion? Cu^2+ + Zn ---->Zn^2 + Cu Given- E for Cu^2/Cu = 0.34 V, E for Zn^2/Zn = -0.76 V, F = 96500 V

I can find out Gibbs free energy from formula,
G=-EnF
But what is the maximum work obtainable?

Answer 1

`DeltaG^@=-212.3"kJ/mol of zinc"`

The maximum work available is `-DeltaG^@` under standard conditions.

Explanation:

If 1 Coulomb of charge is moved through a potential difference of 1 Volt then 1 Joule of work is done.

The charge on a mole of electrons is given by `F` the Faraday Constant.

So:

If `F` Coulombs is moved through a potential difference of `E` Volts then `FE` Joules of work is done.

If `n` is the number of moles of electrons transferred in a cell then it follows that the amount of work done = `nFE`

This is the maximum amount of work available from the cell so:

`w_("available")=nFE`

The connection between this and the free energy change is:

`-DeltaG=w_("available"`

So, under standard conditions:

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

To get `E_(cell)^@` subtract the least positive `E` value from the most positive:

`E_(cell)^@=+0.34-(-0.76)=+1.1"V"`

`:.DeltaG^@=-2xx9.65xx10^4xx1.1=21.23xx10^(4)"J/mol of zinc"`

`DeltaG^@=-212.3"kJ/mol of zinc"`