When $4.5 \, A$ current is passed through $1 \, L$ of $0.6 \, M$ $CuCl_2$ solution for $1.15 \, hour$,calculate the mass of $Cu$ and the change in concentration of the solution. $[Cu = 63.5 \, u]$
(N/A) Time $t = 1.15 \, hour = 1.15 \times 3600 \, s = 4140 \, s$.
Quantity of electricity $Q = I \times t = 4.5 \, A \times 4140 \, s = 18630 \, C$.
Number of moles of electrons $n(e^-) = \frac{Q}{F} = \frac{18630}{96500} \approx 0.19306 \, mol$.
Reaction: $Cu^{2+}_{(aq)} + 2e^- \rightarrow Cu_{(s)}$.
From stoichiometry,$2 \, mol \, e^-$ produce $1 \, mol \, Cu$.
So,moles of $Cu$ deposited $= \frac{0.19306}{2} = 0.09653 \, mol$.
Mass of $Cu = \text{moles} \times \text{atomic mass} = 0.09653 \, mol \times 63.5 \, g/mol \approx 6.13 \, g$.
Change in concentration: Since $1 \, L$ of solution is used,the decrease in $[Cu^{2+}]$ is $0.09653 \, M$.
Quantity of electricity $Q = I \times t = 4.5 \, A \times 4140 \, s = 18630 \, C$.
Number of moles of electrons $n(e^-) = \frac{Q}{F} = \frac{18630}{96500} \approx 0.19306 \, mol$.
Reaction: $Cu^{2+}_{(aq)} + 2e^- \rightarrow Cu_{(s)}$.
From stoichiometry,$2 \, mol \, e^-$ produce $1 \, mol \, Cu$.
So,moles of $Cu$ deposited $= \frac{0.19306}{2} = 0.09653 \, mol$.
Mass of $Cu = \text{moles} \times \text{atomic mass} = 0.09653 \, mol \times 63.5 \, g/mol \approx 6.13 \, g$.
Change in concentration: Since $1 \, L$ of solution is used,the decrease in $[Cu^{2+}]$ is $0.09653 \, M$.
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