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

Question

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

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Answer 1 · 2016-02-04T18:43:42

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

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

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:

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

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

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