For the electrochemical cell,Mg_{(s)} mid Mg^ | vedclass

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(C) The cell reaction is: $Mg_{(s)} + Cu^{2+}_{(aq)} \rightarrow Mg^{2+}_{(aq)} + Cu_{(s)}$The Nernst equation is: $E_{cell} = E^0_{cell} - \frac{RT}{

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$10$

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(C) The cell reaction is: $Mg_{(s)} + Cu^{2+}_{(aq)} ightarrow Mg^{2+}_{(aq)} + Cu_{(s)}$The Nernst equation is: $E_{cell} = E^0_{cell} - \frac{RT}{nF} \ln(Q)$Here,$n = 2$,$E^0_{cell} = 2.70 \ V$,$T = 300 \ K$,and $Q = \frac{[Mg^{2+}]}{[Cu^{2+}]}$.Given $\frac{F}{R} = 11500 \ K \ V^{-1}$,so $\frac{R}{F} = \frac{1}{11500} \ V \ K^{-1}$.Substituting the values into the Nernst equation:$2.67 = 2.70 - \frac{300}{2 	imes 11500} \ln \left( \frac{x}{1} ight)$$-0.03 = -\frac{300}{23000} \ln(x)$$0.03 = \frac{3}{230} \ln(x)$$\ln(x) = 0.03 	imes \frac{230}{3} = 0.01 	imes 230 = 2.30$Since $\ln(10) = 2.30$,we get $x = 10$.

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