Question

For the reduction of silver ions with copper metal, the standard cell potential was found to be +0.46V at 25$^{\circ }C$ . The value of standard Gibbs energy $\Delta {{G}^{\circ }}$will be:
F = 96500 $mol{{e}^{-1}}$
A. -89.0 KJ
B. -79.0 KJ
C. -44.0 KJ
D. -98.0 KJ

Answer 1

Free energy change is a thermochemical quantity and is directly related to the cell potential which is an electrochemical property. Thus, we can find the value of standard Gibbs energy $\Delta {{G}^{\circ }}$ by knowing the value of the cell potential.

Complete step by step answer:
-Electrochemistry is a branch of chemistry that deals with the conversion of chemical energy into electrical energy in electrochemical cells. Process is called electrolysis. When a metal is dipped in electrolyte, potential difference is created between the metal and electrolyte which allows flow of current.
-Reaction takes place at the electrodes. Electrode where oxidation occurs is called anode and electrode where reduction occurs is called cathode. These processes are governed by electrochemical series. The top elements behave as reducing agents and as we go down, the strength of the oxidizing agents increase.

-Energy associated with a chemical reaction which is used to build the potential in galvanic cells is called Gibbs free energy (G). It is given as
$\Delta {{G}^{\circ }}=\Delta {{H}^{\circ }}-T\Delta {{S}^{\circ }}$under standard conditions.
Where $\Delta H\text{=heat energy change and }\Delta \text{S=entropy change}$
-The value of $\Delta {{G}^{\circ }}$is directly related to the cell potential E and can be given as
$\Delta {{G}^{\circ }}=-nFE$
Where n=number of mole of electrons transferred in reaction
F= Faraday’s constant = 96500 C/mol ${{e}^{-}}$
-In the question, we are given the value of F as well as E. If we know the value of n, then we can easily find the value of standard Gibbs energy $\Delta {{G}^{\circ }}$. So, we only need to find the value of n now.

-We can see that we are given the reaction in the question. According to the question, silver ions need to react with copper metal. So we get the value of n from the reaction mentioned in the question. Reaction can be shown as
$\begin{aligned} & Cu+2A{{g}^{+}}\to C{{u}^{2+}}+2Ag \\\ & \Rightarrow \text{n=2} \\\ \end{aligned}$
There are 2 moles of electrons being generated through the reaction and so n=2.

-Now we find the value of standard Gibbs energy $\Delta {{G}^{\circ }}$ as
$\Delta {{G}^{\circ }}=-nFE$
= -2 x 96500 x 0.46
=-89KJ
So, the correct answer is “Option A”.

Note: When metals are present at both the electrodes of the galvanic cell, then which metal will oxidize and which metal will reduce is given by the electrochemical series. It lists all the metals in a sequence of the increasing order of the oxidizing agent. Elements that reduce themselves are oxidizing agents and which oxidize themselves are reducing agents.