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ValentinkaMS [17]
3 years ago
8

Can someone help me solve this? I was given the answer at the end of class but I don't understand how the teacher got it. (Direc

tions: Balance each redox reaction in acid solution using the half reaction method.)
PbO2+I2=>Pb^2+ +IO3^- = I2+8H^+ +5PbO2=>2IO3^- +5Pb^+2 +4H2O
Chemistry
1 answer:
KiRa [710]3 years ago
8 0
Unbalanced chemical reaction: PbO₂ + I₂ → Pb²⁺ +IO₃⁻<span>.
First </span><span>determine oxidation numbers of elements:
1) lead on the left side of reaction has oxidation number +4 (x + 2</span>·(-2) = 0) and on the right side +2, so lead is reduced.
2) iodine has neutal charge (0) on left and oxidation number +5 on the right side of chemical reaction (x + 3 · (-2) = -1), so iodine is oxidized.
3) oxygen has oxidation number -2 on both side of chemical reaction.
Second, write balanced half reactions, because this reaction is acidic solution, add hydrogen ions on one side and water on another side of half reaction; balance atoms and electrons on both half reactions:
Half reaction 1: PbO₂ + 4H⁺ + 2e⁻ → Pb²⁺ + 2H₂O /×5
5PbO₂ + 20H⁺ + 10e⁻ → 5Pb²⁺ + 10H₂O.
Half reaction 2: I₂ + 6H₂O → 2IO₃⁻ + 12H⁺ + 10e⁻.
Sum both half reactions and short same ions: 
5PbO₂ + 20H⁺ + I₂ + 6H₂O → 5Pb²⁺ + 10H₂O + 2IO₃⁻ + 12H⁺.
Abbreviate reaction: 5PbO₂ + 8H⁺ + I₂  → 5Pb²⁺ + 4H₂O + 2IO₃⁻..
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The combustion of ethene in the presence of excess oxygen yields carbon dioxide and water: c2h4 (g) + 3o2 (g) → 2co2 (g) + 2h2o
meriva

Answer:

\boxed{-267.5}

Explanation:

You can calculate the entropy change of a reaction by using the standard molar entropies of reactants and products.

The formula is

\Delta_{r} S^{\circ} = \sum_n {nS_{\text{products}}^{\circ} - \sum_{m} {mS_{\text{reactants}}^{\circ}}}

The equation for the reaction is

                        C₂H₄(g) + 3O₂(g) ⟶ 2CO₂(g) + 2H₂O(ℓ)

ΔS°/J·K⁻¹mol⁻¹   219.5      205.0         213.6         69.9

\Delta_{r} S^{\circ} = (2\times213.6 + 2\times69.9) - (1\times219.5 + 3\times205.0)\\\\= 567.0 - 834.5 = \boxed{-267.5 \text{ J}\cdot\text{K}^{-1} \text{mol}^{-1}}

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Can someone help me here please i need this :(​
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