The E^{circ} for Daniell cell is 1.1 V. Calc | vedclass

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

Vedclass · Accepted Answer

$-212.27 \ kJ$

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(B) The standard Gibbs free energy change is given by the formula: $\Delta G^{\circ} = -nFE^{\circ}_{cell}$.Here,the number of electrons transferred $(n)$ is $2$.The Faraday constant $(F)$ is approximately $96487 \ C \ mol^{-1}$.The standard cell potential $(E^{\circ}_{cell})$ is $1.1 \ V$.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 kilojoules: $\Delta G^{\circ} \approx -212.27 \ kJ \ mol^{-1}$.

The $E^{\circ}$ for Daniell cell is $1.1 \ V$. Calculate $\Delta G^{\circ}$ for the following reaction: $Zn_{(s)} + Cu^{2+}_{(aq)} \rightleftharpoons Zn^{2+}_{(aq)} + Cu_{(s)}$

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For the following cell reaction,$Ag | Ag^{+} | AgCl | Cl^{-} | Cl_2, Pt$
$\Delta G_f^{\circ}(AgCl) = -109 \ kJ/mol$
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$E^{\circ}$ of the cell is

Consider the cell given below:
$Ag_{(s)} | Ag^{\oplus} || Cu^{2+} | Cu_{(s)}$
Given:
$Ag^{\oplus} + e^{-} \to Ag; E^{o} = x$
$Cu^{2+} + 2e^{-} \to Cu; E^{o} = y$
The value of $E^{o}_{cell}$ is:

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