The standard electrode potential for the Daniell cell is $1.1 \, V$. Calculate the standard Gibbs energy for the reaction:
$Zn_{(s)} + Cu^{2+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Cu_{(s)}$
$Zn_{(s)} + Cu^{2+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Cu_{(s)}$
(N/A) The formula for standard Gibbs energy is ${\Delta _r}{G^\Theta } = - nFE_{(cell)}^\Theta $.
In this reaction,the number of electrons transferred $(n)$ is $2$.
The Faraday constant $(F)$ is $96487 \, C \, mol^{-1}$ and the standard cell potential $(E_{(cell)}^\Theta )$ is $1.1 \, V$.
Substituting these values into the equation:
${\Delta _r}{G^\Theta } = - 2 \times 96487 \, C \, mol^{-1} \times 1.1 \, V$
${\Delta _r}{G^\Theta } = - 212271 \, J \, mol^{-1}$
Converting to kilojoules:
${\Delta _r}{G^\Theta } = - 212.27 \, kJ \, mol^{-1}$
In this reaction,the number of electrons transferred $(n)$ is $2$.
The Faraday constant $(F)$ is $96487 \, C \, mol^{-1}$ and the standard cell potential $(E_{(cell)}^\Theta )$ is $1.1 \, V$.
Substituting these values into the equation:
${\Delta _r}{G^\Theta } = - 2 \times 96487 \, C \, mol^{-1} \times 1.1 \, V$
${\Delta _r}{G^\Theta } = - 212271 \, J \, mol^{-1}$
Converting to kilojoules:
${\Delta _r}{G^\Theta } = - 212.27 \, kJ \, mol^{-1}$
Explore More
Similar Questions
The standard oxidation potentials for the half-reactions are given as $Zn \to Zn^{2+} + 2e^{-}; E^o = +0.76 \ V$ and $Fe \to Fe^{2+} + 2e^{-}; E^o = +0.41 \ V$. The $EMF$ for the cell reaction $Fe^{2+} + Zn \to Zn^{2+} + Fe$ is ............ $V$.
MediumIIT 1988
View SolutionStandard electrode potential of three metals $X$,$Y$ and $Z$ are $-1.2 \, V$,$+0.5 \, V$ and $-3.0 \, V$ respectively. The reducing power of these metals will be
MediumAIPMT 2011
View SolutionStandard reduction potentials for the following reactions are :
$Fe^{3+} + 3e^{-} \rightarrow Fe ; E^{\circ}_{1} = -0.036 \ V$
$Fe^{2+} + 2e^{-} \rightarrow Fe ; E^{\circ}_{2} = -0.44 \ V$
What will be the standard electrode potential for
$Fe^{3+} + e^{-} \rightarrow Fe^{2+} (in $V$)?$
$Fe^{3+} + 3e^{-} \rightarrow Fe ; E^{\circ}_{1} = -0.036 \ V$
$Fe^{2+} + 2e^{-} \rightarrow Fe ; E^{\circ}_{2} = -0.44 \ V$
What will be the standard electrode potential for
$Fe^{3+} + e^{-} \rightarrow Fe^{2+} (in $V$)?$
Easy
View SolutionGiven electrode potentials:
$Fe^{3+} + e^- \to Fe^{2+}$; $E^o = 0.771 \, V$
$I_2 + 2e^- \to 2I^-$; $E^o = 0.536 \, V$
$E^o_{\text{cell}}$ for the cell reaction
$2Fe^{3+} + 2I^- \to 2Fe^{2+} + I_2$ is:
$Fe^{3+} + e^- \to Fe^{2+}$; $E^o = 0.771 \, V$
$I_2 + 2e^- \to 2I^-$; $E^o = 0.536 \, V$
$E^o_{\text{cell}}$ for the cell reaction
$2Fe^{3+} + 2I^- \to 2Fe^{2+} + I_2$ is:
Medium
View SolutionThe standard electrode potential is measured by
Easy
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo