# 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?

Feb 4, 2016

$\Delta {G}^{\circ} = - 212.3 \text{kJ/mol of zinc}$

The maximum work available is $- \Delta {G}^{\circ}$ 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 $F E$ 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 = $n F E$

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

${w}_{\text{available}} = n F E$

The connection between this and the free energy change is:

-DeltaG=w_("available"

So, under standard conditions:

$\Delta {G}^{\circ} = - n F {E}_{c e l l}^{\circ}$

To get ${E}_{c e l l}^{\circ}$ subtract the least positive $E$ value from the most positive:

${E}_{c e l l}^{\circ} = + 0.34 - \left(- 0.76\right) = + 1.1 \text{V}$

$\therefore \Delta {G}^{\circ} = - 2 \times 9.65 \times {10}^{4} \times 1.1 = 21.23 \times {10}^{4} \text{J/mol of zinc}$

$\Delta {G}^{\circ} = - 212.3 \text{kJ/mol of zinc}$