Question

Wg of copper deposited in a copper voltameter when an electric current of 2 ampere is passed for 2 hours. If one ampere of electric current is passed for 4 hours in the same voltameter, copper doposited will be

Answer 1

Wg

Answer 2

The problem can be solved using Faraday's first law of electrolysis, which states that the mass of a substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed through the electrolyte.

Mathematically, this is expressed as:

$m = ZIt$

where:

$m$ is the mass of the substance deposited, $Z$ is the electrochemical equivalent of the substance (a constant for a given substance), $I$ is the current, and $t$ is the time for which the current is passed.

Since the same copper voltameter is used, the electrochemical equivalent ($Z$) for copper remains constant.

Let's consider the two cases given:

Case 1:

Mass of copper deposited ($m_1$) = W g Current ($I_1$) = 2 A Time ($t_1$) = 2 hours

Case 2:

Mass of copper deposited ($m_2$) = ? Current ($I_2$) = 1 A Time ($t_2$) = 4 hours

Using Faraday's first law for both cases:

For Case 1:

$W = Z \times I_1 \times t_1$ $W = Z \times 2 \text{ A} \times 2 \text{ hours}$ $W = Z \times 4 \text{ (A-hours)}$ --- (1)

For Case 2:

$m_2 = Z \times I_2 \times t_2$ $m_2 = Z \times 1 \text{ A} \times 4 \text{ hours}$ $m_2 = Z \times 4 \text{ (A-hours)}$ --- (2)

Comparing equation (1) and equation (2), we can see that the right-hand sides are identical.

Therefore,

$m_2 = W$

Alternatively, we can set up a ratio:

$\frac{m_1}{m_2} = \frac{Z I_1 t_1}{Z I_2 t_2}$

Since Z is constant, it cancels out:

$\frac{m_1}{m_2} = \frac{I_1 t_1}{I_2 t_2}$

Substitute the given values:

$\frac{W}{m_2} = \frac{2 \text{ A} \times 2 \text{ hours}}{1 \text{ A} \times 4 \text{ hours}}$ $\frac{W}{m_2} = \frac{4}{4}$ $\frac{W}{m_2} = 1$ $m_2 = W$

The mass of copper deposited in the second case will be W g.