Cu_{(s)} + Sn^{2+}(0.001 M) rightarrow Cu^{2 | vedclass

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(B) The cell reaction is: $Cu_{(s)} + Sn^{2+}(0.001 \ M) \rightarrow Cu^{2+}(0.01 \ M) + Sn_{(s)}$$E^{\ominus}_{cell} = E^{\ominus}_{cathode} - E^{\om

Vedclass · Accepted Answer

$983$

Vedclass · Answer

(B) The cell reaction is: $Cu_{(s)} + Sn^{2+}(0.001 \ M) ightarrow Cu^{2+}(0.01 \ M) + Sn_{(s)}$$E^{\ominus}_{cell} = E^{\ominus}_{cathode} - E^{\ominus}_{anode} = E^{\ominus}_{Sn^{2+}/Sn} - E^{\ominus}_{Cu^{2+}/Cu}$$E^{\ominus}_{cell} = -0.14 \ V - 0.34 \ V = -0.48 \ V$Using the Nernst equation: $E_{cell} = E^{\ominus}_{cell} - \frac{0.0591}{n} \log \frac{[Cu^{2+}]}{[Sn^{2+}]}$$E_{cell} = -0.48 - \frac{0.0591}{2} \log \frac{0.01}{0.001} = -0.48 - 0.02955 	imes \log(10) = -0.48 - 0.02955 = -0.50955 \ V$Gibbs free energy change: $\Delta G = -nFE_{cell}$$\Delta G = -2 	imes 96500 	imes (-0.50955) = 98343.15 \ J \ mol^{-1} = 98.343 \ kJ \ mol^{-1}$Given $\Delta G = x 	imes 10^{-1} \ kJ \ mol^{-1}$,so $x = 983.43$Nearest integer value of $x$ is $983$.

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