Work other than pressure volume work (PΔV) is net useful work. So, total work of a process is,

W = W_{net} + PΔV

W_{net} = W – PΔV

From Gibb’s helmuntz equation,

ΔG = H – TΔS ……………. (i)

We have,

H = E + PΔV

S = Q /T

TΔS = Q

Putting the values of H and TS in eqn (i), we get

ΔG = ΔE + PΔV – Q

From first law of thermodynamics

Q = ΔE + W

Putting the value of Q in above enq we get,

G = Δ E + PΔV – ΔE – W

G = PΔV – W

G = – (W – PΔV)

G = – W_{net}

W_{net} is – ΔG

Thus the net useful work is the amount of decrease in free energy of a system under constant temperature and pressure.

**Relation between free energy change and cell potential**

The net work done by galvanic cell in carrying charge is

W = charge × potential difference = nf E_{cell}

We have,

ΔG = -W_{net}

ΔG = -nfE_{cell}

_{ }**Relation between ****G**** and equilibrium constant**

The variation of G is given by Nernst equation as,

G = ΔG + RT in Q

Where, Q = reaction quotient

Where, the reaction is at equilibrium

g = k (equilibrium constant)

ΔG = O

O = G° + RT in k

ΔG° = -RT InK

ΔG = -2.303 RT logK

The overall reaction for corrosion of iron by oxygen is,

4Fe (s) + 3O2 (g) → 2Fe2O3 (s) Rust

Calculate the free energy change and equilibrium constant for this reaction at 250°c.

Given,

ΔS° and ΔH° = -543 JK-1 and -1652 KJ mol-1

= -1652000 Jmol-1

Δ G° = Δ H° – TΔS°

= 1652000 – 298 × (-543)

= -1490186 J

Again,

ΔG° = -RTInK

= -2.303 RTlogK

1490186 = – 2.303 × 8.314 × 298 × log K

logK = 261.17

K = 10^{261.17}

K = 1.48 × 10261.17

Calculate the free energy at standard condition for galvanic cell having following cell reaction.

2Al (s) + 3 Cu^{ ++} (aq) → 2Al^{+++} (aq) + 3Cu (s)

E° _{Cu++/Cu} = 0.34 V → E° _{Al+++/A1 }= -1.66V

So, the standard cell potential,

E°cell = E°_{cathode} – E°_{anode}

= E°_{cu++/cu} – E°_{Al+++/A1}

= 0.4 – (-1.66)

= 2V

Again, Δ G = -nfEcell

= – 6 × 96500 × 2

= 1.158 × 10^{6}J