For an electrochemical cellSn_{(s)} | Sn^{2+} | vedclass

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(A) The cell reaction is:$Sn_{(s)} + Pb^{2+}_{(aq)} \rightarrow Sn^{2+}_{(aq)} + Pb_{(s)}$$E^{0}_{cell} = E^{0}_{cathode} - E^{0}_{anode} = E^{0}_{Pb^

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$2.15$

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(A) The cell reaction is:$Sn_{(s)} + Pb^{2+}_{(aq)} \rightarrow Sn^{2+}_{(aq)} + Pb_{(s)}$$E^{0}_{cell} = E^{0}_{cathode} - E^{0}_{anode} = E^{0}_{Pb^{2+}/Pb} - E^{0}_{Sn^{2+}/Sn} = -0.13 - (-0.14) = 0.01 \ V$At equilibrium,$E_{cell} = 0$. Using the Nernst equation:$E_{cell} = E^{0}_{cell} - \frac{0.06}{n} \log \frac{[Sn^{2+}]}{[Pb^{2+}]}$Here,$n = 2$ (number of electrons transferred).$0 = 0.01 - \frac{0.06}{2} \log \frac{[Sn^{2+}]}{[Pb^{2+}]}$$0.01 = 0.03 \log \frac{[Sn^{2+}]}{[Pb^{2+}]}$$\log \frac{[Sn^{2+}]}{[Pb^{2+}]} = \frac{0.01}{0.03} = \frac{1}{3} \approx 0.333$$\frac{[Sn^{2+}]}{[Pb^{2+}]} = 10^{0.333} \approx 2.15$

For an electrochemical cell
$Sn_{(s)} | Sn^{2+}(aq, 1 \ M) || Pb^{2+}(aq, 1 \ M) | Pb_{(s)}$
the ratio $\frac{[Sn^{2+}]}{[Pb^{2+}]}$ when this cell attains equilibrium is
(Given $E^{0}_{Sn^{2+}/Sn} = -0.14 \ V$,$E^{0}_{Pb^{2+}/Pb} = -0.13 \ V$,$\frac{2.303 \ RT}{F} = 0.06$)

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For an electrochemical cell$Sn_{(s)} | Sn^{2+}(aq, 1 \ M) || Pb^{2+}(aq, 1 \ M) | Pb_{(s)}$the ratio $\frac{[Sn^{2+}]}{[Pb^{2+}]}$ when this cell attains equilibrium is(Given $E^{0}_{Sn^{2+}/Sn} = -0.14 \ V$,$E^{0}_{Pb^{2+}/Pb} = -0.13 \ V$,$\frac{2.303 \ RT}{F} = 0.06$)