Calculate the time required to deposit 1.98 g of copper from a solution of copper sulfate with a current of 2 A. (1 C deposits 0.0033 g of copper). 2. If the initial mass of the cathode was 5 g and th

From the given information, $1 \text{ C}$ of charge liberates $0.0033 \text{ g}$ of copper. We can set up a proportion:

$$\frac{\text{Mass liberated}}{\text{Charge}} = \text{constant}$$ $$\frac{0.0033 \text{ g}}{1 \text{ C}} = \frac{1.98 \text{ g}}{Q}$$

Solving for $Q$:

$$Q = \frac{1.98 \text{ g} \times 1 \text{ C}}{0.0033 \text{ g}}$$ $$Q = 600 \text{ C}$$

The relationship between charge ($Q$), current ($I$), and time ($t$) is $Q = I \times t$.
We are given $I = 2 \text{ A}$ and we calculated $Q = 600 \text{ C}$.

$$t = \frac{Q}{I}$$ $$t = \frac{600 \text{ C}}{2 \text{ A}}$$ $$t = 300 \text{ s}$$

$$t = 300 \text{ s} \times \frac{1 \text{ minute}}{60 \text{ s}}$$ $$t = 5 \text{ minutes}$$

The correct option is 5 minutes.

$$\boxed{\text{5 minutes}}$$

:

The mass deposited ($m$) is the difference between the final and initial mass of the cathode.

$$m = m_{final} - m_{initial}$$ $$m = 5.01 \text{ g} - 5 \text{ g}$$ $$m = 0.01 \text{ g}$$

According to Faraday's First Law of Electrolysis, the mass deposited ($m$) is given by the formula:

$$m = ZIt$$

Where:

• $m = 0.01 \text{ g}$
• $I = 5 \text{ A}$
• $t = 50 \text{ s}$

Rearrange the formula to solve for $Z$:

$$Z = \frac{m}{It}$$

Substitute the values:

$$Z = \frac{0.01 \text{ g}}{(5 \text{ A}) \times (50 \text{ s})}$$ $$Z = \frac{0.01 \text{ g}}{250 \text{ A s}}$$ $$Z = 0.00004 \text{ g/C}$$

The electrochemical equivalent of the deposited substance is $0.00004 \text{ g/C}$.

$$\boxed{0.00004 \text{ g/C}}$$