To solve the problem of how much metal will be deposited during electrolysis with a different current and time, we can use Faraday's laws of electrolysis. Let's break down the problem step-by-step. ### Step 1: Understand the relationship between current, time, and mass deposited Faraday's first law of electrolysis states that the mass (m) of a substance deposited during electrolysis is directly proportional to the quantity of electricity (Q) passed through the electrolyte. The relationship can be expressed mathematically as: \[ m = k \cdot Q \] where: - \( m \) is the mass of the substance deposited (in grams), - \( k \) is a constant that depends on the substance and its valency, - \( Q \) is the total electric charge passed (in coulombs). The total electric charge \( Q \) can be calculated using the formula: \[ Q = I \cdot t \] where: - \( I \) is the current (in amperes), - \( t \) is the time (in seconds). ### Step 2: Calculate the charge for the first scenario In the first scenario, we have: - Current \( I_1 = 0.5 \, \text{A} \) - Time \( t_1 = 3 \, \text{h} = 3 \times 3600 \, \text{s} = 10800 \, \text{s} \) Now, we can calculate the total charge \( Q_1 \): \[ Q_1 = I_1 \cdot t_1 = 0.5 \, \text{A} \cdot 10800 \, \text{s} = 5400 \, \text{C} \] ### Step 3: Determine the mass deposited in the first scenario According to the problem, this charge deposits 2g of metal. Therefore, we can express this relationship as: \[ 2 \, \text{g} = k \cdot 5400 \, \text{C} \] From this, we can find the constant \( k \): \[ k = \frac{2 \, \text{g}}{5400 \, \text{C}} \] ### Step 4: Calculate the charge for the second scenario In the second scenario, we have: - Current \( I_2 = 1.5 \, \text{A} \) - Time \( t_2 = 1 \, \text{h} = 3600 \, \text{s} \) Now, we can calculate the total charge \( Q_2 \): \[ Q_2 = I_2 \cdot t_2 = 1.5 \, \text{A} \cdot 3600 \, \text{s} = 5400 \, \text{C} \] ### Step 5: Determine the mass deposited in the second scenario Now we can use the same constant \( k \) to find the mass deposited in the second scenario: Using the relationship \( m = k \cdot Q_2 \): Since we already calculated \( Q_2 = 5400 \, \text{C} \) and we know \( k = \frac{2 \, \text{g}}{5400 \, \text{C}} \): \[ m = \left(\frac{2 \, \text{g}}{5400 \, \text{C}}\right) \cdot 5400 \, \text{C} = 2 \, \text{g} \] ### Conclusion: Final Answer Thus, the quantity of the same metal that would be deposited by the current of 1.5 A flowing for 1 hour is **2g**. ### Explanation of Other Options - **Option B (6g)**: This option assumes that the mass deposited would be three times the original amount due to the increase in current. However, the time is also a factor, and in this case, the total charge remains the same. - **Option C (10g)**: This option suggests an even larger increase in mass, which is incorrect as it does not take into account the time factor correctly. - **Option D (18g)**: This option is also incorrect as it assumes a linear increase without considering the time duration properly. ### Revision Summary - Faraday's laws of electrolysis relate mass deposited to the charge passed. - The charge \( Q \) is calculated using \( Q = I \cdot t \). - The mass deposited is directly proportional to the charge passed. - In this case, the mass deposited remains the same (2g) despite the increase in current because the total charge is unchanged.