To solve the problem of how much mass of a univalent metal will be deposited when a current of \(2I\) is passed through an electrolyte solution for 10 minutes, we need to apply Faraday's laws of electrolysis. Let's break this down step-by-step. ### Step 1: Understanding Faraday's Laws of Electrolysis Faraday's first law states that the mass of a substance deposited or liberated at an electrode during electrolysis is directly proportional to the quantity of electricity (charge) that passes through the electrolyte. Mathematically, this can be expressed as: \[ m = k \cdot Q \] Where: - \(m\) is the mass of the substance deposited, - \(k\) is a constant that depends on the substance (for a univalent metal, \(k\) is related to its molar mass and the charge), - \(Q\) is the total electric charge passed through the electrolyte, given by \(Q = I \cdot t\), where \(I\) is the current and \(t\) is the time in seconds. ### Step 2: Calculate the Charge for the Initial Condition In the initial scenario, a current \(I\) is passed for 40 minutes. First, we need to convert the time into seconds: \[ t_1 = 40 \text{ minutes} \times 60 \text{ seconds/minute} = 2400 \text{ seconds} \] Now, we can calculate the total charge \(Q_1\) that passed through the solution: \[ Q_1 = I \cdot t_1 = I \cdot 2400 \text{ seconds} \] According to Faraday's first law, the mass \(X\) of the metal deposited is: \[ X = k \cdot Q_1 = k \cdot (I \cdot 2400) \] ### Step 3: Calculate the Charge for the New Condition Now, we need to find out how much mass will be deposited when a current of \(2I\) is passed for 10 minutes. Again, we convert the time into seconds: \[ t_2 = 10 \text{ minutes} \times 60 \text{ seconds/minute} = 600 \text{ seconds} \] Now, we calculate the total charge \(Q_2\) for this new condition: \[ Q_2 = 2I \cdot t_2 = 2I \cdot 600 \text{ seconds} = 1200I \] ### Step 4: Calculate the Mass Deposited in the New Condition Using Faraday's first law again, the mass \(m\) of the metal deposited when the current \(2I\) is passed for 10 minutes is: \[ m = k \cdot Q_2 = k \cdot (1200I) \] ### Step 5: Relate the Two Masses Now, we can relate the two masses \(X\) and \(m\): From the first condition, we have: \[ X = k \cdot (2400I) \] From the second condition, we have: \[ m = k \cdot (1200I) \] To find the relationship between \(m\) and \(X\), we can express \(m\) in terms of \(X\): \[ m = \frac{1200I}{2400I} \cdot X = \frac{1200}{2400} \cdot X = \frac{1}{2}X \] ### Conclusion: Final Answer Thus, the mass of the metal deposited when a current \(2I\) is passed through the solution for 10 minutes is: \[ m = \frac{X}{2} \text{ g} \] The correct option is **B. \(X/2\) g**. ### Explanation of Other Options - **A. \(X/4\) g**: This option suggests that the mass deposited is a quarter of \(X\). This is incorrect because the current is doubled, and the time is reduced, leading to a larger mass than \(X/4\). - **C. \(2X\) g**: This option suggests that the mass deposited is double \(X\). This is incorrect because while the current is doubled, the time is also significantly reduced, which does not lead to a doubling of the mass. - **D. \(4X\) g**: This option suggests that the mass deposited is four times \(X\). This is incorrect as it does not take into account the reduced time of 10 minutes. ### Revision Summary - Faraday's first law states that mass deposited is proportional to the charge passed. - Charge \(Q\) is calculated as \(Q = I \cdot t\). - For the first condition, \(X = k \cdot (I \cdot 2400)\). - For the second condition, \(m = k \cdot (2I \cdot 600)\). - The relationship shows that \(m = \frac{X}{2}\), leading to the correct answer of \(X/2\) g.