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Following reaction takes place in the cell:
$Zn(s)\ +\ A{{g}_{2}}O(s)\ +\ {{H}_{2}}O(l)\ \to \ \ Z{{n}^{2+}}+\ Ag(s)\ +\ 2O{{H}^{-}}(aq)$
Calculate ${{\Delta }_{r}}{{G}^{0}}$ of the reaction.
[Given: $E_{(Z{{n}^{2+}}/Zn)}^{0}$ = -0.76 V, $E_{(A{{g}^{+}}/Ag)}^{0}$= 0.80 V, 1F = 96,500 $C\ mo{{l}^{-1}}$]

Last updated date: 14th Jun 2024
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Hint: In the above cell reaction, identify the element that undergoes oxidation and the element that undergoes reduction. Based on this you can determine the electrodes, anode and cathode. Now you can apply the formula to calculate the emf of the cell. The emf of the cell must be substituted in the Gibbs-Helmholtz equation to calculate ${{\Delta }_{r}}{{G}^{0}}$ of the above cell reaction.

Complete step-by-step answer:
-Electrochemistry is the branch of physical chemistry that mainly deals with the relationship between electricity and identifiable chemical change.
-Electrochemical reaction is a chemical reaction in which current is externally supplied or produced through a spontaneous chemical reaction.
-Chemical reactions where electrons are directly transferred between the constituent molecules or atoms are called oxidation-reduction or rather redox reactions.
-In general, we can say that a reaction is spontaneous when the emf of the cell is positive. It indicates that the cell reaction is feasible and thus occurs.
-The formula for EMF is:
EMF = ${{E}^{0}}_{cathode}-{{E}^{0}}_{anode}$
-Where, the potential values are taken in standard electrode potential.
-Oxidation takes place at anode and reduction at cathode. So, Zn is at anode and silver at cathode. We will now substitute the electrode potentials of zinc and silver in the formula given above.
EMF = 0.80 - (-0.76) V
EMF = 1.56 V
-The Gibbs-Helmholtz equation gives a relation between the electrochemical factor and thermodynamic factor of the cell. The equation is given below:
${{\Delta }_{r}}{{G}^{0}}=-nF{{E}^{0}}$
${{\Delta }_{r}}{{G}^{0}}$ is Gibbs free energy for the reaction,
n is the number of moles of electrons involved in the reaction,
F is the amount of charge present in 1 mole of electrons
${{E}^{0}}$ is the emf of cell
The above reaction involves exchange of 2 mole of electrons. The value of F is given to us as 96500 F. We will now substitute these values in the question above.
${{\Delta }_{r}}{{G}^{0}}=-(2).(96500).(1.56)\ J$ = - 301 kJ

Therefore, the ${{\Delta }_{r}}{{G}^{0}}$ of the reaction is calculated as -301 kJ.

Note: Gibbs free energy mentioned in the above formula is a measure of unwanted work that happens in a reaction. This is the reason why change in Gibbs energy must be negative for the overall reaction to take place and be thermodynamically feasible.