(D) The formula for standard Gibbs energy is $\Delta G^{\circ} = -nFE^{\circ}_{\text{cell}}$.Here,$n = 2$ (number of electrons transferred),$F = 96500

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(D) The formula for standard Gibbs energy is $\Delta G^{\circ} = -nFE^{\circ}_{\text{cell}}$.Here,$n = 2$ (number of electrons transferred),$F = 96500 \ C \ mol^{-1}$ (Faraday constant),$E^{\circ}_{\text{cell}} = 1.1 \ V$.Substituting the values:$\Delta G^{\circ} = -2 \times 96500 \times 1.1 \ J \ mol^{-1}$.$\Delta G^{\circ} = -212300 \ J \ mol^{-1}$.Converting to $kJ \ mol^{-1}$:$\Delta G^{\circ} = -212.3 \ kJ \ mol^{-1}$.

Class 12 Chemistry Electrochemistry Electrode potential and ECell

Medium

The standard electrode potential for a $Daniell$ cell is $1.1 \ V$. Calculate the standard Gibbs energy for the reaction: $[Zn_{(s)} + Cu^{2+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Cu_{(s)}]$

A
$-200.3 \ kJ \ mol^{-1}$
B
$-199.4 \ kJ \ mol^{-1}$
C
$-202.3 \ kJ \ mol^{-1}$
D
$-212.3 \ kJ \ mol^{-1}$

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Electrochemistry

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Electrode potential and ECell

On the basis of the following $E^o$ values,the strongest oxidizing agent is:
$[Fe(CN)_6]^{4-} \to [Fe(CN)_6]^{3-} + e^-; E^o = -0.35 \ V$
$Fe^{2+} \to Fe^{3+} + e^-; E^o = -0.77 \ V$

MediumAIPMT 2008

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Given $E^{0}_{Fe^{2+}|Fe} = -0.441 \, V$ and $E^{0}_{Fe^{3+}|Fe^{2+}} = 0.771 \, V$,calculate the standard $EMF$ for the reaction $Fe + 2Fe^{3+} \rightarrow 3Fe^{2+}$ in $V$.

Difficult

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$FeO_4^{2-}$ $\xrightarrow{2.0 \ V} Fe^{3+}$ $\xrightarrow{0.8 \ V} Fe^{2+}$ $\xrightarrow{-0.44 \ V} Fe^0$
In the above diagram,the standard electrode potentials are given in volts. The value of $E_{FeO_4^{2-}/Fe^{2+}}^{\Theta}$ is: (in $V$)

DifficultJEE MAIN 2025

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The $E^{0}_{Red}$ values of $A, B, C,$ and $D$ are $0.8 \, V, 0.79 \, V, 0.34 \, V,$ and $-2.37 \, V$ respectively. Which element can displace the other three from their salt solutions?

Medium

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For the cell reaction,$2Ce^{4+} + Co \to 2Ce^{3+} + Co^{2+}$,$E_{cell}^o$ is $1.89 \, V$ and $E_{Co/Co^{2+}}^o = +0.28 \, V$. If $E_{Ce^{4+}/Ce^{3+}}^o = x$,then $x$ is ............. $V$.

Medium

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The standard electrode potential for a $Daniell$ cell is $1.1 \ V$. Calculate the standard Gibbs energy for the reaction: $[Zn_{(s)} + Cu^{2+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Cu_{(s)}]$

Similar Questions

On the basis of the following $E^o$ values,the strongest oxidizing agent is:$[Fe(CN)_6]^{4-} \to [Fe(CN)_6]^{3-} + e^-; E^o = -0.35 \ V$$Fe^{2+} \to Fe^{3+} + e^-; E^o = -0.77 \ V$

Given $E^{0}_{Fe^{2+}|Fe} = -0.441 \, V$ and $E^{0}_{Fe^{3+}|Fe^{2+}} = 0.771 \, V$,calculate the standard $EMF$ for the reaction $Fe + 2Fe^{3+} \rightarrow 3Fe^{2+}$ in $V$.

$FeO_4^{2-}$ $\xrightarrow{2.0 \ V} Fe^{3+}$ $\xrightarrow{0.8 \ V} Fe^{2+}$ $\xrightarrow{-0.44 \ V} Fe^0$In the above diagram,the standard electrode potentials are given in volts. The value of $E_{FeO_4^{2-}/Fe^{2+}}^{\Theta}$ is: (in $V$)

The $E^{0}_{Red}$ values of $A, B, C,$ and $D$ are $0.8 \, V, 0.79 \, V, 0.34 \, V,$ and $-2.37 \, V$ respectively. Which element can displace the other three from their salt solutions?

For the cell reaction,$2Ce^{4+} + Co \to 2Ce^{3+} + Co^{2+}$,$E_{cell}^o$ is $1.89 \, V$ and $E_{Co/Co^{2+}}^o = +0.28 \, V$. If $E_{Ce^{4+}/Ce^{3+}}^o = x$,then $x$ is ............. $V$.

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On the basis of the following $E^o$ values,the strongest oxidizing agent is:
$[Fe(CN)_6]^{4-} \to [Fe(CN)_6]^{3-} + e^-; E^o = -0.35 \ V$
$Fe^{2+} \to Fe^{3+} + e^-; E^o = -0.77 \ V$

$FeO_4^{2-}$ $\xrightarrow{2.0 \ V} Fe^{3+}$ $\xrightarrow{0.8 \ V} Fe^{2+}$ $\xrightarrow{-0.44 \ V} Fe^0$
In the above diagram,the standard electrode potentials are given in volts. The value of $E_{FeO_4^{2-}/Fe^{2+}}^{\Theta}$ is: (in $V$)