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3 years ago

How is the potential (voltage) of a redox reaction determined using the reduction

Chemistry

1 answer:

LuckyWell [14K]3 years ago

6 0

Answer:

caused by the ability of electrons to flow from one half cell too the other

Explanation:

How is the potential voltage of a redox reaction?

The potential difference is caused by the ability of electrons to flow from one half cell to the other. Electrons are able to move between electrodes because the chemical reaction is a redox reaction. A redox reaction occurs when a certain substance is oxidized, while another is reduced.

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) In the Deacon process for the manufacture of chlorine, HCl and O2 react to form Cl2 and H2O. Sufficient air (21 mole% O2, 79%

LiRa [457]

Answer:

Here's what I get.

Explanation:

1. Write the chemical equation

$\rm 4HCl + O$_{2} \longrightarrow \,$ 2Cl$_{2}$ + 2H$_{2}$O$

Assume that we start with 4 L of HCl

2. Calculate the theoretical volume of oxygen

$\text{V}_{\text{O}_{2}}= \text{4 L HCl} \times \dfrac{\text{1 L O}_{2}}{\text{4 L HCl}} = \text{1 L O}_{2}}$

3. Add 35% excess

$\text{V}_{\text{O}_{2}}= \text{1 L O}_{2}} \times 1.35 = \text{1.35 L O}_{2}}$

4. Calculate the theoretical volume of nitrogen

$\text{V}_{\text{N}_{2}} = \text{1.35 L O}_{2}} \times \dfrac{\text{79 L N}_{2}}{\text{21 L O}_{2}}} = \text{5.08 L N}_{2}}$

4. Calculate volumes of reactant used up

Only 85 % of the HCl is converted.

We can summarize the volumes in an ICE table

4HCl + O₂ + N₂ → 2Cl₂ + 2H₂O

I/L: 4 1.35 5.08 0 0

C/L: -0.85(4) -0.85(1) 0 +0.85(2) +0.85(2)

E/L: 0.60 0.50 5.08 1.70 1.70

5. Calculate the mole fractions of each gas in the product stream

Total volume = (0.60 + 0.50 + 5.08 + 1.70 + 1.70) L = 9.58 L

$\chi = \dfrac{\text{V}_{\text{component}}}{\text{V}_\text{total}} = \dfrac{\text{ V}_{\text{component}}}{\text{9.58}} = \text{0.1044V}_{\text{component}}\\\\\chi_{\text{HCl}} = 0.1044\times 0.60 = 0.063\\\\\chi_{\text{O}_{2}} = 0.1044\times 0.50 = 0.052\\\\\chi_{\text{N}_{2}} = 0.1044\times 5.08 = 0.530\\\\\chi_{\text{Cl}_{2}} = 0.1044\times 1.70 = 0.177\\\\\chi_{\text{H$_{2}${O}}} = 0.1044\times 1.70 = 0.177\\\\$

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