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:
${\Delta _r}{G^\Theta } = - 212.27 \, kJ \, mol^{-1}$
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Similar Questions
For the cell involving the following reaction: $Zn_{(s)} + Ni^{2+}_{(aq)} \longrightarrow Zn^{2+}_{(aq)} + Ni_{(s)}$. Given $E^{\circ}_{\text{cell}} = 0.5 \ V$. What is the standard Gibbs energy change of the cell reaction (in $kJ$)?
$A$ galvanic cell with electrode potential of $A = + 2.23 \ V$ and $B = - 1.43 \ V$. The value of $E^o_{cell}$ is .............. $V$.
Medium
View SolutionCan $E_{cell}^o$ or $\Delta_r G^o$ for a cell reaction ever be equal to zero?
Medium
View SolutionGiven the standard reduction potentials at $25\,^oC$ for the following half-reactions:
$Zn^{2+}_{(aq)} + 2e^- \rightleftharpoons Zn_{(s)}, E^o = -0.76\,V$
$Cr^{3+}_{(aq)} + 3e^- \rightleftharpoons Cr_{(s)}, E^o = -0.74\,V$
$2H^+_{(aq)} + 2e^- \rightleftharpoons H_{2(g)}, E^o = 0.00\,V$
$Fe^{3+}_{(aq)} + e^- \rightleftharpoons Fe^{2+}_{(aq)}, E^o = +0.77\,V$
Which of the following is the strongest reducing agent?
$Zn^{2+}_{(aq)} + 2e^- \rightleftharpoons Zn_{(s)}, E^o = -0.76\,V$
$Cr^{3+}_{(aq)} + 3e^- \rightleftharpoons Cr_{(s)}, E^o = -0.74\,V$
$2H^+_{(aq)} + 2e^- \rightleftharpoons H_{2(g)}, E^o = 0.00\,V$
$Fe^{3+}_{(aq)} + e^- \rightleftharpoons Fe^{2+}_{(aq)}, E^o = +0.77\,V$
Which of the following is the strongest reducing agent?
Easy
View SolutionWhich one among the following is the strongest reducing agent?
$Fe^{2+} + 2e^- \to Fe \ (E^\circ = -0.44 \ V)$
$Ni^{2+} + 2e^- \to Ni \ (E^\circ = -0.25 \ V)$
$Sn^{2+} + 2e^- \to Sn \ (E^\circ = -0.14 \ V)$
$Fe^{3+} + e^- \to Fe^{2+} \ (E^\circ = +0.77 \ V)$
$Fe^{2+} + 2e^- \to Fe \ (E^\circ = -0.44 \ V)$
$Ni^{2+} + 2e^- \to Ni \ (E^\circ = -0.25 \ V)$
$Sn^{2+} + 2e^- \to Sn \ (E^\circ = -0.14 \ V)$
$Fe^{3+} + e^- \to Fe^{2+} \ (E^\circ = +0.77 \ V)$
Medium
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