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Question: What is \( \Delta G \) , if \( n = 2 \) and cell potential is \( 2.226{\text{ V}} \) ?...

What is ΔG\Delta G , if n=2n = 2 and cell potential is 2.226 V2.226{\text{ V}} ?

Explanation

Solution

Hint : ΔG\Delta G is the change in the free energy. It is a state function which determines whether a reaction is favorable or not. The relation between the Gibbs energy and the cell potential is ΔG=−nFE(cell)\Delta G = - nF{E_{(cell)}} This equation of Gibbs free energy is used only for redox reactions so ΔG\Delta G gives the spontaneity of redox reactions.

Complete Step By Step Answer:
The Gibbs free energy (G) of a system is a measure of the amount of usable energy (energy that can do work) in that system. ΔG\Delta G Is the change in the Gibbs free energy of a system from initial state to the final state. It can be positive or negative according to whether the reaction is spontaneous or nonspontaneous. For thermodynamics ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S
In electrochemical reactions, ΔG\Delta G is the change in free energy which is equal to the electrochemical potential also called as cell potential times the electrical charge q (equals to nF) transferred in a redox reaction. The equation of free energy change is given by:
ΔG=−nFE(cell)\Delta G = - nF{E_{(cell)}} where ΔG\Delta G is the change in Gibbs free energy, n is the number of electrons transferred , F is Faraday’s constant having value 96485J/Vmol{96485_{}}J/{V_{}}mol (for calculations it is approximately taken as 96500J/Vmol{96500_{}}J/{V_{}}mol )and E(cell){E_{(cell)}} is the cell potential.
It is asked in the question to find out ΔG\Delta G ; we are going to use the above formula:
⇒\Rightarrow ΔG=−nFE(cell)\Delta G = - nF{E_{(cell)}}
Given that the number of electrons transferred ‘n’ =2= 2 and cell potential E(cell)=2.226V{E_{(cell)}} = {2.226_{}}V
Thus ⇒\Rightarrow ΔG=−nFE(cell)\Delta G = - nF{E_{(cell)}}
ΔG=−2×96500×2.226\Delta G = - 2 \times 96500 \times 2.226
ΔG=−2×96500×2.226\Delta G = - 2 \times 96500 \times 2.226
ΔG=−429618\Delta G = - 429618
⇒\Rightarrow ΔG=−42.9618Kjmol−1\Delta G = - {42.9618_{}}Kjmo{l^{ - 1}}
Therefore the Gibbs free energy change ΔG=−42.9618Kjmol−1\Delta G = - {42.9618_{}}Kjmo{l^{ - 1}} .

Note :
Under standard experimental conditions the formula ΔG=−nFE(cell)\Delta G = - nF{E_{(cell)}} changes to ΔG0=−nFE0(cell)\Delta {G^0} = - nF{E^0}_{(cell)} where ΔG0\Delta {G^0} is the standard Gibbs free energy and E0(cell){E^0}_{(cell)} is standard cell potential. This usually happens when the reactants and the products are combining at the standard conditions of temperature and pressure.