Question

For the following cell reaction the standard Gibbs energy change will be Mg(s) + Zn2*(aq) → Mg2*(aq) + Zn(s)

(Given E° Mg2-g = -2.36 V, E Zn2+iZn =-0.76V)

(a) -308.8 kJ mol-1
(b) -275.6 kJ mol-1
(c ) -45.2 kJ mol-1
(d) -75.35 kJ mol-1​

Answer 1

Given:

for the reaction, Mg(s) + Zn2*(aq) → Mg2*(aq) + Zn(s)

E(Mg2+/Mg) = -2.36 V  = at anode

E(Zn2+/Zn) = -0.76 V    = at cathode

To Find:

the standard Gibbs energy =?

Solution:

Now, for the standard E of the cell, we know that

E(cell) = E(cathode) - E(anode)

    on putting the given values

E(cell) = -0.76 - (-2.36)

         = 1.6V

We also know that the expression for standard Gibbs energy,

ΔG = -nFE(cell)

now we know n = 2 (2 electrons change involved)

F(faraday constant) = 96485 C/mol and value of E(cell)

On putting values,

ΔG = - 2 × 96485 × 1.6 = -308752 J/mol = -308.8kJ/mol

Thus, the correct option is (a)  -308.8kJ/mol