Calculate the volume of oxygen evolved at STP during the electrolysis of acidified water with a current of 2.5 A for 1.5 minutes.

To calculate the volume of oxygen evolved, we need to follow these steps:

1. Write the balanced half-reaction for the evolution of oxygen.
2. Calculate the total charge passed through the solution.
3. Determine the moles of electrons transferred.
4. Calculate the moles of oxygen produced using stoichiometry.
5. Use the ideal gas law to find the volume of oxygen at the given conditions.

Step 1: Write the half-reaction for oxygen evolution.
When acidified water undergoes electrolysis, oxygen gas is produced at the anode. The half-reaction is:

$$2\text{H}_2\text{O}(l) \rightarrow \text{O}_2(g) + 4\text{H}^+(aq) + 4\text{e}^-$$

This equation shows that 4 moles of electrons are required to produce 1 mole of oxygen gas.

Step 2: Calculate the total charge passed ($Q$).
Given current, $I = 2.5 \text{ A}$
Given time, $t = 1.5 \text{ mins}$
First, convert time to seconds:

$$t = 1.5 \text{ mins} \times 60 \text{ s/min} = 90 \text{ s}$$

Now, calculate the charge using the formula $Q = I \times t$:

$$Q = 2.5 \text{ A} \times 90 \text{ s} = 225 \text{ C}$$

Step 3: Determine the moles of electrons transferred ($n_e$).
Given Faraday's constant, $F = 96500 \text{ C/mol e}^-$
The moles of electrons transferred are:

$$n_e = \frac{Q}{F} = \frac{225 \text{ C}}{96500 \text{ C/mol e}^-} \approx 0.0023316 \text{ mol e}^-$$

Step 4: Calculate the moles of oxygen produced ($n_{\text{O}_2}$).
From the half-reaction, 4 moles of electrons produce 1 mole of $\text{O}_2$.

$$n_{\text{O}_2} = n_e \times \frac{1 \text{ mol O}_2}{4 \text{ mol e}^-} = 0.0023316 \text{ mol e}^- \times \frac{1}{4} \approx 0.0005829 \text{ mol O}_2$$

Step 5: Use the ideal gas law to find the volume of oxygen ($V$).
The ideal gas law is $PV = nRT$.
We need to solve for $V$: $V = \frac{nRT}{P}$.
Given conditions:
Temperature, $T = 285 \text{ K}$
Pressure, $P = 0.91 \times 10^5 \text{ Nm}^{-2}$ (which is Pascals, Pa)
Moles of $\text{O}_2$, $n = 0.0005829 \text{ mol}$
The ideal gas constant, $R = 8.314 \text{ J mol}^{-1} \text{ K}^{-1}$ (or $\text{Pa m}^3 \text{ mol}^{-1} \text{ K}^{-1}$)

Substitute the values into the ideal gas law:

$$V = \frac{(0.0005829 \text{ mol}) \times (8.314 \text{ J mol}^{-1} \text{ K}^{-1}) \times (285 \text{ K})}{0.91 \times 10^5 \text{ Pa}}$$ $$V = \frac{1.3807086}{91000} \text{ m}^3$$ $$V \approx 0.0000151726 \text{ m}^3$$

Convert the volume from cubic meters to cubic centimeters:

$$1 \text{ m}^3 = (100 \text{ cm})^3 = 10^6 \text{ cm}^3$$ $$$$