The standard electrode potential for the Dani | vedclass

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(C) The standard Gibbs energy change $\Delta G^{\circ}$ is given by the formula: $\Delta G^{\circ} = -nFE^{\circ}_{cell}$.Here,the number of electrons

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$-212.27 \ kJ \ mol^{-1}$

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(C) The standard Gibbs energy change $\Delta G^{\circ}$ is given by the formula: $\Delta G^{\circ} = -nFE^{\circ}_{cell}$.Here,the number of electrons transferred in the reaction is $n = 2$.The standard electrode potential is $E^{\circ}_{cell} = 1.1 \ V$.The Faraday constant is $F = 96487 \ C \ mol^{-1}$.Substituting these values: $\Delta G^{\circ} = -(2) 	imes (96487 \ C \ mol^{-1}) 	imes (1.1 \ V)$.$\Delta G^{\circ} = -212271.4 \ J \ mol^{-1}$.Converting to $kJ \ mol^{-1}$: $\Delta G^{\circ} = -212.27 \ kJ \ mol^{-1}$.

The standard electrode potential for the Daniell cell is $1.1 \ V$. What will be the value of standard Gibbs energy for the reaction?
$Zn_{(s)} + Cu_{(aq)}^{2+} \rightarrow Zn_{(aq)}^{2+} + Cu_{(s)}$
$(1 \ F = 96487 \ C \ mol^{-1})$

Similar Questions

$A$ galvanic cell consists of a copper electrode and a standard hydrogen electrode. If $E^\circ (Cu^{2+}_{(aq)} | Cu_{(s)}) = +0.34 \text{ V}$, identify the reaction taking place at the positive electrode during the working of the cell.

The e.m.f. of a cell whose half cells are given below is .............. $V$
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Then for $Cu^{2+} + e^- \to Cu^{+}$ ; $E^o$ will be ?

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The standard electrode potential for the Daniell cell is $1.1 \ V$. What will be the value of standard Gibbs energy for the reaction?$Zn_{(s)} + Cu_{(aq)}^{2+} \rightarrow Zn_{(aq)}^{2+} + Cu_{(s)}$$(1 \ F = 96487 \ C \ mol^{-1})$