Given the following half-cell reaction; 2Cl_(aq) → CI_2(g) + 2e, how many mole of electrons will be required to produce 1.12dm³ of chlorine gas at s.t.p? (Molar volume of a gas at s.t.p. = 22.4dm ³)

tionwyne

1 year ago

To determine how many moles of electrons are required to produce \( 1.12 \, \text{dm}^3 \) of chlorine gas at standard temperature and pressure (s.t.p.), we can follow these steps:

### Step 1: Understand the given half-cell reaction
The half-cell reaction provided is:
\[ 2Cl^-(aq) \rightarrow Cl_2(g) + 2e^- \]

This indicates that **2 moles of electrons (\( 2e^- \))** are required to produce **1 mole of chlorine gas (\( Cl_2 \))**.

### Step 2: Calculate the moles of chlorine gas produced
The molar volume of a gas at s.t.p. is \( 22.4 \, \text{dm}^3 \). To find the moles of \( Cl_2 \) in \( 1.12 \, \text{dm}^3 \):
\[
\text{Moles of } Cl_2 = \frac{\text{Volume of } Cl_2}{\text{Molar volume}} = \frac{1.12 \, \text{dm}^3}{22.4 \, \text{dm}^3/\text{mol}} = 0.05 \, \text{mol}
\]

### Step 3: Relate moles of \( Cl_2 \) to moles of electrons
From the half-cell reaction, **1 mole of \( Cl_2 \)** requires **2 moles of electrons**. Therefore, for \( 0.05 \, \text{mol} \) of \( Cl_2 \):
\[
\text{Moles of electrons} = 2 \times \text{Moles of } Cl_2 = 2 \times 0.05 \, \text{mol} = 0.10 \, \text{mol}
\]

### Final Answer
The number of moles of electrons required is:
\[
\boxed{0.10}
\]

Eb3n3z3r

8 months ago

oh its correct

Ayomipo2603

5 years ago

can someone pls explain

Eb3n3z3r

8 months ago

the answer is wrong