(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

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

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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Electrode potential and ECell

Consider the reaction $M_{(aq)}^{n+} + n e^{-} \to M_{(s)}$. The standard reduction potential values of the elements $M_1$,$M_2$,and $M_3$ are $-0.34 \ V$,$-3.05 \ V$,and $-1.66 \ V$ respectively. The order of their reducing power will be

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If the electrode potential of $Zn^{2+} | Zn$ is $-0.76 \, V$ and the potential of $Cu^{2+} | Cu$ is $0.34 \, V$,what will be the $EMF$ of the cell formed by both electrodes in $V$?

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Given that the standard reduction potentials for $M^{+}/M$ and $N^{+}/N$ electrodes at $298 \ K$ are $0.52 \ V$ and $0.25 \ V$ respectively. Which of the following is correct in respect of the following electrochemical cell?
$M | M^{+} || N^{+} | N$

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Given $E_{Fe^{3+}|Fe}^0 = -0.036 \ V$ and $E_{Fe^{2+}|Fe}^0 = -0.439 \ V$,calculate $E^0_{cell}$ for the reaction $Fe^{3+} + e^{-} \rightarrow Fe^{2+}$.

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Calculate $E_{\text{cell}}^{\circ}$ for $Cd_{(s)}|Cd^{2+}_{(1M)}||Ag^{+}_{(1M)}|Ag_{(s)}$. Given: $E^{\circ}_{Cd^{2+}/Cd} = -0.403 \ V$ and $E^{\circ}_{Ag^{+}/Ag} = 0.799 \ V$. (in $V$)

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

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Consider the reaction $M_{(aq)}^{n+} + n e^{-} \to M_{(s)}$. The standard reduction potential values of the elements $M_1$,$M_2$,and $M_3$ are $-0.34 \ V$,$-3.05 \ V$,and $-1.66 \ V$ respectively. The order of their reducing power will be

If the electrode potential of $Zn^{2+} | Zn$ is $-0.76 \, V$ and the potential of $Cu^{2+} | Cu$ is $0.34 \, V$,what will be the $EMF$ of the cell formed by both electrodes in $V$?

Given that the standard reduction potentials for $M^{+}/M$ and $N^{+}/N$ electrodes at $298 \ K$ are $0.52 \ V$ and $0.25 \ V$ respectively. Which of the following is correct in respect of the following electrochemical cell?$M | M^{+} || N^{+} | N$

Given $E_{Fe^{3+}|Fe}^0 = -0.036 \ V$ and $E_{Fe^{2+}|Fe}^0 = -0.439 \ V$,calculate $E^0_{cell}$ for the reaction $Fe^{3+} + e^{-} \rightarrow Fe^{2+}$.

Calculate $E_{\text{cell}}^{\circ}$ for $Cd_{(s)}|Cd^{2+}_{(1M)}||Ag^{+}_{(1M)}|Ag_{(s)}$. Given: $E^{\circ}_{Cd^{2+}/Cd} = -0.403 \ V$ and $E^{\circ}_{Ag^{+}/Ag} = 0.799 \ V$. (in $V$)

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Given that the standard reduction potentials for $M^{+}/M$ and $N^{+}/N$ electrodes at $298 \ K$ are $0.52 \ V$ and $0.25 \ V$ respectively. Which of the following is correct in respect of the following electrochemical cell?
$M | M^{+} || N^{+} | N$