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A certain current liberates 0.5g of hyd...

A certain current liberates `0.5g` of hydrogen in 2 hours. How many grams of copper can be liberated by the same current flowing for the same time in a copper sulphate solution ?

A

`12.7g`

B

`15.9g`

C

`31.8g`

D

`63.5g`

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The correct Answer is:
To solve the problem of how many grams of copper can be liberated by the same current flowing for the same time in a copper sulfate solution, we can use Faraday's second law of electrolysis. Here’s a step-by-step solution: ### Step 1: Understand the relationship between the substances liberated According to Faraday's second law of electrolysis, the weight of a substance liberated at an electrode during electrolysis is directly proportional to its equivalent weight when the same quantity of electricity is passed through different electrolytic solutions. ### Step 2: Set up the equation Let: - \( W_H \) = weight of hydrogen liberated = 0.5 g - \( W_C \) = weight of copper deposited (what we want to find) - \( E_H \) = equivalent weight of hydrogen - \( E_C \) = equivalent weight of copper From Faraday's second law, we have: \[ \frac{W_H}{E_H} = \frac{W_C}{E_C} \] ### Step 3: Calculate the equivalent weights 1. **Equivalent weight of hydrogen (H)**: - Molecular weight of hydrogen (H₂) = 2 g/mol - Equivalent weight of hydrogen = \(\frac{Molar\ weight}{n} = \frac{2}{2} = 1\) g/equiv 2. **Equivalent weight of copper (Cu)**: - Atomic weight of copper = 63.5 g/mol - For copper, it typically forms Cu²⁺ ions, so \( n = 2 \). - Equivalent weight of copper = \(\frac{63.5}{2} = 31.75\) g/equiv ### Step 4: Substitute the values into the equation Now substituting the known values into the equation: \[ \frac{0.5}{1} = \frac{W_C}{31.75} \] ### Step 5: Solve for \( W_C \) Cross-multiplying gives: \[ W_C = 0.5 \times 31.75 \] Calculating this: \[ W_C = 15.875 \text{ g} \] ### Step 6: Round to the appropriate significant figures Rounding 15.875 g gives approximately 15.9 g. ### Conclusion Thus, the weight of copper that can be liberated by the same current flowing for the same time in a copper sulfate solution is approximately **15.9 grams**.

To solve the problem of how many grams of copper can be liberated by the same current flowing for the same time in a copper sulfate solution, we can use Faraday's second law of electrolysis. Here’s a step-by-step solution: ### Step 1: Understand the relationship between the substances liberated According to Faraday's second law of electrolysis, the weight of a substance liberated at an electrode during electrolysis is directly proportional to its equivalent weight when the same quantity of electricity is passed through different electrolytic solutions. ### Step 2: Set up the equation Let: - \( W_H \) = weight of hydrogen liberated = 0.5 g ...
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