To solve the problem of determining the current required to produce 1 mole of \( S_2O_8^{2-} \) per hour through the electrolysis of \( HSO_4^{-1} \) ions with a current efficiency of 75%, we can follow these steps:
### Step 1: Understand the Electrolysis Reaction
The electrolysis of \( HSO_4^{-1} \) ions produces \( S_2O_8^{2-} \) ions according to the reaction:
\[
2 HSO_4^{-1} \rightarrow S_2O_8^{2-} + 2 e^{-}
\]
This means that for the production of 1 mole of \( S_2O_8^{2-} \), 2 moles of electrons are required.
### Step 2: Calculate the Charge Required
To find the total charge (\( Q \)) required to produce 1 mole of \( S_2O_8^{2-} \), we can use Faraday's law of electrolysis. The charge required to transfer 1 mole of electrons is given by Faraday's constant, which is approximately \( 96500 \, C/mol \).
Since 2 moles of electrons are needed:
\[
Q = n \times F = 2 \, \text{moles} \times 96500 \, C/mol = 193000 \, C
\]
### Step 3: Calculate the Time in Seconds
We need to convert the time from hours to seconds. Since we want to produce 1 mole in 1 hour:
\[
t = 1 \, \text{hour} = 60 \, \text{minutes} \times 60 \, \text{seconds/minute} = 3600 \, \text{seconds}
\]
### Step 4: Calculate the Current without Efficiency
Using the formula \( Q = I \times t \), we can rearrange to find the current (\( I \)):
\[
I = \frac{Q}{t} = \frac{193000 \, C}{3600 \, s} \approx 53.61 \, A
\]
### Step 5: Adjust for Current Efficiency
Since the current efficiency is given as 75%, we need to adjust our current calculation:
\[
\text{Actual Current} = \frac{I}{\text{Efficiency}} = \frac{53.61 \, A}{0.75} \approx 71.48 \, A
\]
### Step 6: Round the Answer
Finally, we round the answer to two decimal places:
\[
I \approx 71.50 \, A
\]
Thus, the current that should be employed to achieve a production rate of 1 mole of \( S_2O_8^{2-} \) per hour, considering the current efficiency, is approximately **71.50 A**.
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