Electrochemistry - Electrolytic Cells and Electrolysis

A solution of $CuSO_4$ is electrolyzed for $10$ minutes with a current of $1.5$ amperes. What is the mass of copper deposited at the cathode? (Atomic mass of $Cu = 63.5\ u$)

Given: $I = 1.5\ A$, $t = 10 \times 60 = 600\ s$. Charge $Q = I \times t = 1.5 \times 600 = 900\ C$. The cathode reaction is $Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$. To deposit $1\ mol$ of $Cu$, $2\ F$ ($2 \times 96500\ C$) is required. Mass $w = \frac{M \times Q}{n \times F} = \frac{63.5 \times 900}{2 \times 96500} = 0.296\ g$.

We first calculate the total charge $Q$ in Coulombs. Using the stoichiometry of the reduction reaction ($n=2$), we apply Faraday's Law to find the mass deposited.

During the electrolysis of molten $Al_2O_3$, how many Coulombs are required to produce $40.5\ g$ of $Al$? (Atomic mass of $Al = 27\ u$)

The reaction at the cathode is $Al^{3+} + 3e^- \rightarrow Al(s)$. Moles of $Al = \frac{40.5}{27} = 1.5\ mol$. Since $1\ mol$ of $Al$ requires $3\ F$ of charge, $1.5\ mol$ requires $1.5 \times 3\ F = 4.5\ F$. Total charge $Q = 4.5 \times 96500 = 434250\ C$.

The number of moles of electrons required is calculated based on the valency of Aluminum ($n=3$). Multiplying the moles of substance by $n$ and Faraday's constant gives the total charge.

A current of $2.0\ A$ is passed through a solution of silver nitrate ($AgNO_3$) for $30$ minutes. Calculate the mass of silver deposited at the cathode. (Atomic mass of $Ag = 108\ u$, $1\ F = 96500\ C\ mol^{-1}$)

1. Calculate total charge ($Q$): $Q = I \times t$ $Q = 2.0\ A \times (30 \times 60)\ s = 3600\ C$
2. Write the cathode reaction: $Ag^+(aq) + e^- \rightarrow Ag(s)$ This shows $1\ mol$ of $e^-$ ($96500\ C$) deposits $108\ g$ of $Ag$.
3. Calculate mass ($w$): $w = \frac{108 \times 3600}{96500}$ $w \approx 4.03\ g$

The quantity of electricity is found by multiplying current and time in seconds. Using Faraday's constant and the stoichiometry of the silver ion reduction, the mass is derived.

Two cells containing $ZnSO_4$ and $CuSO_4$ are connected in series. If $0.654\ g$ of Zinc is deposited in the first cell, calculate the mass of Copper deposited in the second cell. (Atomic mass: $Zn = 65.4\ u$, $Cu = 63.5\ u$)

According to Faraday's Second Law: $\frac{w_1}{w_2} = \frac{E_1}{E_2}$

1. Calculate Equivalent Weights ($E = \frac{\text{Molar Mass}}{n}$): $E_{Zn} = \frac{65.4}{2} = 32.7$ $E_{Cu} = \frac{63.5}{2} = 31.75$
2. Substitute values: $\frac{0.654}{w_{Cu}} = \frac{32.7}{31.75}$ $w_{Cu} = \frac{0.654 \times 31.75}{32.7}$ $w_{Cu} = 0.635\ g$

Since the cells are in series, the same charge passes through both. The ratio of the masses of zinc and copper deposited equals the ratio of their chemical equivalent weights.