Calculate E_{cell}^{circ} if the equilibrium | vedclass

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(A) The relationship between standard cell potential and equilibrium constant is given by the formula: $E_{cell}^{\circ} = \frac{0.0592}{n} \log_{10}

Vedclass · Accepted Answer

$0.36$

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(A) The relationship between standard cell potential and equilibrium constant is given by the formula: $E_{cell}^{\circ} = \frac{0.0592}{n} \log_{10} K$ at $298 \ K$.The given reaction is $2 Cu_{(aq)}^{+} \longrightarrow Cu_{(aq)}^{2+} + Cu_{(s)}$.The half-reactions are:Reduction: $Cu_{(aq)}^{+} + e^{-} \longrightarrow Cu_{(s)}$Oxidation: $Cu_{(aq)}^{+} \longrightarrow Cu_{(aq)}^{2+} + e^{-}$Thus,the number of electrons transferred,$n = 1$.Substituting the values: $E_{cell}^{\circ} = \frac{0.0592}{1} \log_{10} (1.2 	imes 10^6)$.$E_{cell}^{\circ} = 0.0592 (\log 1.2 + \log 10^6)$.$E_{cell}^{\circ} = 0.0592 (0.079 + 6)$.$E_{cell}^{\circ} = 0.0592 	imes 6.079 \approx 0.36 \ V$.

Calculate $E_{cell}^{\circ}$ if the equilibrium constant for the following reaction is $1.2 \times 10^6$.
$2 Cu_{(aq)}^{+} \longrightarrow Cu_{(aq)}^{2+} + Cu_{(s)}$ (in $V$)

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Calculate $E_{cell}^{\circ}$ if the equilibrium constant for the following reaction is $1.2 \times 10^6$.$2 Cu_{(aq)}^{+} \longrightarrow Cu_{(aq)}^{2+} + Cu_{(s)}$ (in $V$)