1 F electricity was passed through Cu^{2+} ( | vedclass

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(A) $1$. For $Cu^{2+} + 2e^{-} \longrightarrow Cu$:Initial moles of $Cu^{2+} = 1.5 \ M 	imes 1 \ L = 1.5 \ mol$.$1 \ F$ electricity corresponds to $1

Vedclass · Accepted Answer

$0.40$

Vedclass · Answer

(A) $1$. For $Cu^{2+} + 2e^{-} \longrightarrow Cu$:Initial moles of $Cu^{2+} = 1.5 \ M 	imes 1 \ L = 1.5 \ mol$.$1 \ F$ electricity corresponds to $1 \ mol$ of electrons.Since $2 \ mol$ of electrons reduce $1 \ mol$ of $Cu^{2+}$,$1 \ mol$ of electrons reduces $0.5 \ mol$ of $Cu^{2+}$.Remaining moles of $Cu^{2+} = 1.5 - 0.5 = 1.0 \ mol$.$[Cu^{2+}] = 1.0 \ M$.$2$. For $Ag^{+} + e^{-} \longrightarrow Ag$:Initial moles of $Ag^{+} = 0.2 \ M 	imes 1 \ L = 0.2 \ mol$.$0.1 \ F$ electricity corresponds to $0.1 \ mol$ of electrons.$1 \ mol$ of electrons reduces $1 \ mol$ of $Ag^{+}$,so $0.1 \ mol$ of electrons reduces $0.1 \ mol$ of $Ag^{+}$.Remaining moles of $Ag^{+} = 0.2 - 0.1 = 0.1 \ mol$.$[Ag^{+}] = 0.1 \ M$.$3$. Cell reaction: $Cu(s) + 2Ag^{+}(aq) \longrightarrow Cu^{2+}(aq) + 2Ag(s)$.$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 0.8 - 0.34 = 0.46 \ V$.Using Nernst equation: $E = E^{\circ}_{cell} - \frac{0.06}{n} \log \frac{[Cu^{2+}]}{[Ag^{+}]^2}$.$E = 0.46 - \frac{0.06}{2} \log \frac{1}{(0.1)^2} = 0.46 - 0.03 \log(100) = 0.46 - 0.03 	imes 2 = 0.46 - 0.06 = 0.40 \ V$.

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