Find the ratio of electrochemical equivalents of silver and aluminium .

Answer

To find the ratio of electrochemical equivalents of silver (Ag) and aluminum (Al), we can follow these steps:

### Step 1: Understand the Definition
Electrochemical equivalent (E) is defined as the weight of an electrolyte deposited by passing one coulomb of charge.

### Step 2: Determine the Electrochemical Equivalent of Silver
For silver (Ag):
- The half-reaction is: \( \text{Ag}^+ + e^- \rightarrow \text{Ag} \)
- This means 1 mole of Ag is deposited by 1 mole of electrons (1 Faraday = 96500 coulombs).
- The atomic mass of silver (Ag) is 108 g/mol.

Using the formula for electrochemical equivalent:
\[ E_{Ag} = \frac{\text{Atomic mass of Ag}}{\text{Faraday constant}} = \frac{108 \, \text{g/mol}}{96500 \, \text{C/mol}} \]

### Step 3: Determine the Electrochemical Equivalent of Aluminum
For aluminum (Al):
- The half-reaction is: \( \text{Al}^{3+} + 3e^- \rightarrow \text{Al} \)
- This means 1 mole of Al is deposited by 3 moles of electrons (3 Faradays = 3 × 96500 coulombs).
- The atomic mass of aluminum (Al) is 27 g/mol.

Using the formula for electrochemical equivalent:
\[ E_{Al} = \frac{\text{Atomic mass of Al}}{3 \times \text{Faraday constant}} = \frac{27 \, \text{g/mol}}{3 \times 96500 \, \text{C/mol}} \]

### Step 4: Calculate the Ratio of Electrochemical Equivalents
Now, we can find the ratio of the electrochemical equivalents of silver and aluminum:
\[ \text{Ratio} = \frac{E_{Ag}}{E_{Al}} = \frac{\frac{108}{96500}}{\frac{27}{3 \times 96500}} \]

### Step 5: Simplify the Ratio
- Cancel out \( 96500 \) from the numerator and denominator:
\[ \text{Ratio} = \frac{108}{\frac{27}{3}} = \frac{108 \times 3}{27} \]
- Simplifying further:
\[ \text{Ratio} = \frac{324}{27} = 12 \]

### Final Answer
The ratio of electrochemical equivalents of silver to aluminum is:
\[ \text{Ratio} = 12 : 1 \]

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