0.50 , L of a 0.60 , M CuSO_4 solution is ele | vedclass

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Question

(D) The reaction at the cathode is: $Cu^{2+} (aq) + 2e^- ightarrow Cu (s)$.Initial moles of $Cu^{2+} = Molarity 	imes Volume = 0.60 \, M 	imes 0.5

Vedclass · Accepted Answer

$0.514$

Vedclass · Answer

(D) The reaction at the cathode is: $Cu^{2+} (aq) + 2e^- ightarrow Cu (s)$.Initial moles of $Cu^{2+} = Molarity 	imes Volume = 0.60 \, M 	imes 0.50 \, L = 0.30 \, mol$.Charge passed $(Q)$ = $I 	imes t = 4.60 \, A 	imes (30.0 	imes 60) \, s = 8280 \, C$.Moles of electrons passed = $\frac{Q}{F} = \frac{8280}{96500} \approx 0.0858 \, mol$.Since $2 \, mol$ of electrons deposit $1 \, mol$ of $Cu$,moles of $Cu^{2+}$ reduced = $\frac{0.0858}{2} = 0.0429 \, mol$.Moles of $Cu^{2+}$ remaining = $0.30 - 0.0429 = 0.2571 \, mol$.Final concentration $[Cu^{2+}] = \frac{0.2571 \, mol}{0.50 \, L} = 0.5142 \, M \approx 0.514 \, M$.

$0.50 \, L$ of a $0.60 \, M$ $CuSO_4$ solution is electrolyzed for a period of $30.0 \, min$ using a current of $4.60 \, A$. If inert electrodes are used,what is the final concentration of $Cu^{2+}$ remaining in the solution? $M$

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