The equilibrium constant for the reaction Zn_ | vedclass

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(C) The cell reaction is $Zn_{(s)} + Sn^{2+}_{(aq)} \rightleftharpoons Zn^{2+}_{(aq)} + Sn_{(s)}$.For this reaction,the number of electrons transferre

Vedclass · Accepted Answer

$17$

Vedclass · Answer

(C) The cell reaction is $Zn_{(s)} + Sn^{2+}_{(aq)} ightleftharpoons Zn^{2+}_{(aq)} + Sn_{(s)}$.For this reaction,the number of electrons transferred is $n = 2$.The relationship between standard cell potential and equilibrium constant is given by $E_{cell}^0 = \frac{0.059}{n} \log_{10} K_{eq}$.Substituting the values: $E_{cell}^0 = \frac{0.059}{2} \log_{10} (1 	imes 10^{20}) = \frac{0.059 	imes 20}{2} = 0.59 \ V$.We know that $E_{cell}^0 = E_{cathode}^0 - E_{anode}^0 = E_{Sn^{2+}/Sn}^0 - E_{Zn^{2+}/Zn}^0$.$0.59 \ V = E_{Sn^{2+}/Sn}^0 - (-0.76 \ V)$.$E_{Sn^{2+}/Sn}^0 = 0.59 - 0.76 = -0.17 \ V$.The standard electrode potential for $Sn/Sn^{2+}$ is the negative of $E_{Sn^{2+}/Sn}^0$,so $E_{Sn/Sn^{2+}}^0 = -(-0.17 \ V) = 0.17 \ V$.$0.17 \ V = 17 	imes 10^{-2} \ V$.Thus,the magnitude is $17$.

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