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Gibbs Free Energy of Reaction MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Quick Facts

  • Feasibility and Gibbs Free Energy of Reaction is considered one the most difficult concept.

  • 27 Questions around this concept.

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QuestionEASY

If the \mathrm{E^o_{cell}} for a given reaction has a negative value, which of the following gives the correct relationships for the values of \mathrm{\Delta G^ \circ} and Keq?

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For the reduction of silver ions with copper metal, the standard cell potential was found to be +0.46 V at 25°C. The value of standard Gibbs energy, \mathrm{\Delta G^{\circ}} will be (F=96500 C mol-1)

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The electrode potentials for

\mathrm{Cu^{2+}_{(aq)}+e^{-}\rightarrow Cu^{+}_{(aq)}}

and \mathrm{Cu^{+}_{(aq)}+e^{-}\rightarrow Cu_{(s)}}

are +0.15 V and +0.50, respectively. The value of \mathrm{E^o_{Cu^{2+}/Cu}} will be:

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The values of \Delta H \ and \ \Delta S for the reaction, C_{\left ( graphite \right )}+CO_{2}\ _{\left ( g \right )}\rightarrow 2CO_{\left ( g \right )}  are 170 kJ and 170JK^{-1}, respectively. The reaction will be spontaneous at

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Given

E_{Cl_{2}/Cl^{-1}}^{0}=1.36V,E_{Cr^{3 +}/Cr}^{0}=-0.74V

E_{Cr_{2}O_{7}^{2-}/Cr^{3+}}^{0}=1.33V,E_{MnO_{4}^{-}/Mn^{2+}}^{0}=1.1V

Among the following, the strongest reducing agent is :

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Given below are two statements: one is labeled as Assertion A and the other is labeled as Reason R :
Assertion A: In equation \Delta G=-n F E_{\text {cell }} , value of \Delta G depends on n.
Reasons  R : E_{\text {cell }} is an intensive property and \Delta G  is an extensive property.

In the light of the above statements, choose the correct answer from the options given below :

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Concepts Covered - 1

Feasibility and Gibbs Free Energy of Reaction

Let n faraday charge be taken out of a cell of emf E, then work done by the cell will be calculated as:
Work = Charge × Potential
Work done by the cell is equal to the decrease in the free energy.

-\Delta \mathrm{G}=\mathrm{nFE}

Similarly, maximum obtainable work from the cell at standard condition will be:

\mathrm{W}_{\max }=\mathrm{nF} \mathrm{E}_{\mathrm{cell}}^{0} \quad \text { where } \mathrm{E}_{\mathrm{cell}}^{0}=\text { standard emf of standard cell potential }

\mathrm{-\Delta G^{\circ}=n F E_{\text {cell }}^{0}}

Variation of E.M.F. (Ecell) with Temperature
The temperature coefficient of e.m.f. of the cell is written as:

\Delta \mathrm{G}=\Delta \mathrm{H}+\mathrm{T}\left(\frac{\mathrm{d}(\Delta \mathrm{G})}{\mathrm{dT}}\right)_{\mathrm{Const.} \text {Pressure }}

\Rightarrow \quad-\mathrm{nFE}_{\mathrm{cell}}=\Delta \mathrm{H}+\mathrm{T}\left(\frac{\mathrm{d}\left(-\mathrm{nFE}_{\mathrm{cell}}\right)}{\mathrm{dT}}\right)_{\mathrm{P}} \quad\quad\quad\left[\because \Delta \mathrm{G}=-\mathrm{nFE}_{\mathrm{cell}}\right]

\Rightarrow \quad\left(\frac{\mathrm{dE}_{\mathrm{cell}}}{\mathrm{dT}}\right)_{\mathrm{P}}=\frac{\Delta \mathrm{H}}{\mathrm{nFT}}+\frac{\mathrm{E}_{\mathrm{cell}}}{\mathrm{T}}

Enthalpy Change: From above equation, we have:

\Delta \mathrm{H}=\Delta \mathrm{G}-\mathrm{T}\left(\frac{\mathrm{d}}{\mathrm{dT}}(\Delta \mathrm{G})\right)_{\mathrm{P}}

Thus, enthalpy(ΔH) is given as:

\Delta \mathrm{H}=-\mathrm{nF}\left[\mathrm{E}_{\mathrm{cell}}-\mathrm{T}\left(\frac{\mathrm{dE}_{\mathrm{cell}}}{\mathrm{dT}}\right)_{\mathrm{P}}\right]

Entropy change: For entropy change we have:

\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}

\Delta \mathrm{S}=-\left(\frac{\mathrm{d}}{\mathrm{dT}}(\Delta \mathrm{G})\right)_{\mathrm{P}}=\mathrm{nF}\left(\frac{\mathrm{dE}_{\mathrm{cell}}}{\mathrm{dT}}\right)_{\mathrm{P}}

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Feasibility and Gibbs Free Energy of Reaction

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