Chemistry, 01.04.2020 07:53 zalyndevola

What will be the end product for each electrode after electrolysis if the solution is concentrated aqueous sodium chloride ?

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Answer from: loveket143

Cathode: Hydrogen gas

Anode: Chlorine gas

Explanation:

① Write down the ions present in the electrolyte

Cations: Na⁺, H⁺

Anions: Cl⁻ , OH⁻

② Decide which ions are preferentially discharged.

These are the factors:

For the discharge of cations,

- Reactivity series

(The lower the position of the cation in the reactivity series, the easier it is to be discharged)

For the discharge of anions,

- Concentration effect (look at this first)

(The more concentrated the ion, the easier for it to be discharged)

- If solution is not concentrated (dilute), look at the position of the anion on the electrochemical series.

(The lower the position of the anion on the electrochemical series, the easier of it to be discharged)

In this case:

For cations, H⁺ ions are selectively discharged at the cathode as its position is lower than Na⁺ in the reactivity series.

Anion: Cl⁻ ions, being more concentrated, are selectively discharged at the anode.

☆For electrolysis,

Cation at the cathode (-ve terminal)

Anion at the anode (+ve terminal)

In summary, here's what happened at each electrode:

Cathode

- H⁺ selectively discharged

- ionic half equation: 2H⁺ (aq) +2e⁻ → H₂ (g)

Anode

- Cl⁻ ions selectively discharged

- ionic half equation: 2Cl⁻ (aq) → Cl₂ (g) + 2e⁻

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

Answer from: Quest

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