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?

1 Answer
Feb 4, 2016

Answer:

#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"#