How many coulombs must be applied to a cell for the electrolytic production of 245 g `NaClO_(4)` from `NaClO_(3)`. The anode efficiency for the desired reaction is 60%.

Answer

To determine how many coulombs must be applied to a cell for the electrolytic production of 245 g of NaClO₄ from NaClO₃ with an anode efficiency of 60%, we can follow these steps:

### Step 1: Determine the molar mass of NaClO₄
The molar mass of NaClO₄ can be calculated as follows:
- Sodium (Na): 22.99 g/mol
- Chlorine (Cl): 35.45 g/mol
- Oxygen (O): 16.00 g/mol × 4 = 64.00 g/mol

Adding these together:
\[ \text{Molar mass of NaClO₄} = 22.99 + 35.45 + 64.00 = 122.44 \text{ g/mol} \]

### Step 2: Calculate the number of moles of NaClO₄ produced
Using the molar mass, we can find the number of moles of NaClO₄ produced from 245 g:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{245 \text{ g}}{122.44 \text{ g/mol}} \approx 2.00 \text{ moles} \]

### Step 3: Write the balanced chemical equation
The balanced chemical equation for the reaction can be represented as:
\[ \text{NaClO}_3 + \text{H}_2\text{O} \rightarrow \text{NaClO}_4 + \text{H}_2 \]
From this equation, we see that 1 mole of NaClO₃ produces 1 mole of NaClO₄.

### Step 4: Determine the total charge required for the reaction
To find the total charge required, we need to know the number of electrons transferred in the reaction. The formation of NaClO₄ from NaClO₃ involves the transfer of 2 electrons (as per the oxidation state change of chlorine).

Using Faraday's law, the charge (Q) can be calculated using:
\[ Q = n \times F \]
Where:
- \( n \) = number of moles of electrons transferred
- \( F \) = Faraday's constant (approximately 96485 C/mol)

Since we have 2 moles of NaClO₄ produced, we need:
\[ n = 2 \text{ moles of NaClO₄} \times 2 \text{ moles of electrons/mole of NaClO₄} = 4 \text{ moles of electrons} \]
Now, substituting into the equation:
\[ Q = 4 \text{ moles} \times 96485 \text{ C/mol} = 385940 \text{ C} \]

### Step 5: Adjust for anode efficiency
Since the anode efficiency is 60%, we need to adjust the total charge:
\[ \text{Actual charge required} = \frac{Q}{\text{Efficiency}} = \frac{385940 \text{ C}}{0.60} \approx 643233.33 \text{ C} \]

### Final Answer
The total charge that must be applied to the cell for the electrolytic production of 245 g of NaClO₄ from NaClO₃ with 60% anode efficiency is approximately **643233.33 C**.

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