In an electroplating experiment, mg of silver is deposited when 4 A of current flows for 2 min. The amount of silver (in g) deposited by 6 A of current for 40 s will be

Options

A. 4m
B. `(m)/(2)`
C. 2m
D. `(m)/(4)`

Answer

To solve the problem of how much silver is deposited when 6 A of current flows for 40 seconds, we can use the first law of electrolysis, which states that the mass (m) of a substance deposited is directly proportional to the current (i) and the time (t) for which the current flows. The formula is given by:

\[ m = z \cdot i \cdot t \]

where:
- \( m \) = mass of the substance deposited (in grams)
- \( z \) = electrochemical equivalent (a constant for the substance)
- \( i \) = current (in amperes)
- \( t \) = time (in seconds)

### Step 1: Calculate the mass of silver deposited with the initial conditions.

Given:
- Current \( i_1 = 4 \, \text{A} \)
- Time \( t_1 = 2 \, \text{min} = 2 \times 60 \, \text{s} = 120 \, \text{s} \)
- Mass deposited \( m_1 = m \, \text{g} \) (unknown mass)

Using the formula:
\[ m_1 = z \cdot i_1 \cdot t_1 \]
\[ m_1 = z \cdot 4 \cdot 120 \]

### Step 2: Calculate the mass of silver deposited with the new conditions.

Given:
- Current \( i_2 = 6 \, \text{A} \)
- Time \( t_2 = 40 \, \text{s} \)

Using the formula:
\[ m_2 = z \cdot i_2 \cdot t_2 \]
\[ m_2 = z \cdot 6 \cdot 40 \]

### Step 3: Set up the ratio of the masses deposited.

From the first law of electrolysis, we can set up the ratio:
\[ \frac{m_2}{m_1} = \frac{z \cdot i_2 \cdot t_2}{z \cdot i_1 \cdot t_1} \]

Substituting the values:
\[ \frac{m_2}{m} = \frac{6 \cdot 40}{4 \cdot 120} \]

### Step 4: Simplify the ratio.

Calculating the right-hand side:
\[ \frac{m_2}{m} = \frac{240}{480} = \frac{1}{2} \]

### Step 5: Solve for \( m_2 \).

From the ratio, we have:
\[ m_2 = \frac{m}{2} \]

### Conclusion:

Thus, the amount of silver deposited by 6 A of current for 40 seconds is:
\[ m_2 = \frac{m}{2} \]

### Final Answer:
The amount of silver deposited will be \( \frac{m}{2} \) grams.

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