Complete and balance the following equation: NO2−(aq)+Cr2O72−(aq)→Cr3+(aq)+NO3−(aq)(acidic solution) Express your answer as a ne

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Complete and balance the following equation: NO2−(aq)+Cr2O72−(aq)→Cr3+(aq)+NO3−(aq)(acidic solution) Express your answer as a ne

t chemical equation including phases. Request Answer Part B Complete and balance the following equation: S(s)+HNO3(aq)→H2SO3(aq)+N2O(g)(acidic solution) Express your answer as a net chemical equation including phases. nothing Request Answer Part C Complete and balance the following equation: Cr2O72−(aq)+CH3OH(aq)→HCO2H(aq)+Cr3+(aq)(acidic solution) Express your answer as a net chemical equation including phases. nothing Request Answer Part D Complete and balance the following equation: BrO3−(aq)+N2H4(aq)→Br2(l)+N2(g)(acidic solution) Express your answer as a net chemical equation including phases. nothing Request Answer Part E Complete and balance the following equation: NO2−(aq)+Al(s)→NH4+(aq)+AlO2−(aq)(basic solution) Express your answer as a net chemical equation including phases. nothing Request Answer Part F Complete and balance the following equation: H2O2(aq)+ClO2(aq)→ClO2−(aq)+O2(g)(basic solution) Express your answer as a net chemical equation including phases. nothing

Chemistry

1 answer:

Georgia [21] · Answer 1 · 2021-12-08T23:05:46+03:00

Answer:

Part A: Cr₂O₇²⁻(aq) + 8H+ + 3NO₂⁻(aq) → 2Cr³⁺(aq) + 3NO₃⁻(aq) + 4H₂O.

Part B: 2HNO₃(aq) + 2S(s) + H₂O(l) → N₂O(g) + 2H₂SO₃(aq).

Part C: 2Cr₂O₇²⁻(aq) + 3CH₃OH(aq) + 16H⁺ → 4Cr³⁺(aq) + 3HCO₂H(aq) + 11H₂O.

Part D: 4BrO₃⁻(aq) + 5N₂H₄(aq) + 4H⁺(aq) → 2Br₂(I) + 5N₂(g) + 12H₂O(aq).

Part E: NO₂⁻(aq) + 2Al(s) + OH⁻(aq) + H₂O(aq) → NH₃(aq) + 2AlO₂⁻(aq).

Part F: H₂O₂(aq) + ClO₂(aq) → ClO₂⁻(aq) + O₂(g) + 2H⁺(aq).

Explanation:

Part A: Complete and balance the following equation: NO₂⁻(aq) + Cr₂O₇²⁻(aq) → Cr³⁺(aq) + NO₃⁻(aq) (acidic solution). Express your answer as a net chemical equation including phases.

To balance and write the net chemical equation, we should write the two-half reactions:

The two half reactions are:

The reduction reaction: Cr₂O₇²⁻(aq) + 14H⁺ + 6e → 2Cr³⁺(aq) + 7H₂O.

The oxidation reaction: H₂O(aq) + NO₂⁻(aq) → NO₃⁻(aq) + 2H⁺(aq) + 2e.

Before adding the reactions number of electron gained must be equal to number of electrons lost. we multiply the reduction reaction by 1 (be the same) and the oxidation reaction by 3 (3H₂O(aq) + 3NO₂⁻(aq) → 3NO₃⁻(aq) + 6H⁺(aq) + 6e) to equalize the no. of electrons in the two-half reactions.

Add up both reactions:

Cr₂O₇²⁻(aq) + 14H⁺ + 6e → 2Cr³⁺(aq) + 7H₂O.

3H₂O(aq) + 3NO₂⁻(aq) → 3NO₃⁻(aq) + 6H⁺(aq) + 6e.

So, the net redox reaction will be:

Cr₂O₇²⁻(aq) + 8H⁺ + 3NO₂⁻(aq) → 2Cr³⁺(aq) + 3NO₃⁻(aq) + 4H₂O.

Part B: Complete and balance the following equation: S(s) + HNO₃(aq) → H₂SO₃(aq) + N₂O(g) (acidic solution) Express your answer as a net chemical equation including phases.

To balance and write the net chemical equation, we should write the two-half reactions:

The two half reactions are:

The oxidation reaction: S(s) + 3H₂O(l) → H₂SO₃(aq) + 4H⁺(aq) + 4e.

The reduction reaction: 2HNO₃(aq) + 8H⁺(aq) + 8e → N₂O(g) + 5H₂O(l).

Before adding the reactions number of electron gained must be equal to number of electrons lost. we multiply the oxidation reaction by 2 (2S(s) + 6H₂O(l) → 2H₂SO₃(aq) + 8H⁺(aq) + 8e) and the reduction reaction by 1 (be the same) to equalize the no. of electrons in the two-half reactions.

Add up both reactions:

2S(s) + 6H₂O(l) → 2H₂SO₃(aq) + 8H⁺(aq) + 8e.

2HNO₃(aq) + 8H⁺(aq) + 8e → N₂O(g) + 5H₂O(l).

So, the net redox reaction will be:

2HNO₃(aq) + 2S(s) + H₂O(l) → N₂O(g) + 2H₂SO₃(aq).

Part C: Complete and balance the following equation: Cr₂O₇²⁻(aq) + CH₃OH(aq) → HCO₂H(aq) + Cr³⁺(aq) (acidic solution), Express your answer as a net chemical equation including phases.

To balance and write the net chemical equation, we should write the two-half reactions:

The two half reactions are:

The reduction reaction: Cr₂O₇²⁻(aq) + 14H⁺ + 6e → 2Cr³⁺(aq) + 7H₂O.

The oxidation reaction: CH₃OH(aq) + H₂O(aq) → HCO₂H(aq) + 4H⁺(aq) + 4e.

Before adding the reactions number of electron gained must be equal to number of electrons lost. we multiply the reduction reaction by 2 (2Cr₂O₇²⁻(aq) + 28H⁺ + 12e → 4Cr³⁺(aq) + 14H₂O) and the reduction reaction by 3 (3CH₃OH(aq) + 3H₂O(aq) → 3HCO₂H(aq) + 12H⁺(aq) + 12e) to equalize the no. of electrons in the two-half reactions.

Add up both reactions:

2Cr₂O₇²⁻(aq) + 28H⁺ + 12e → 4Cr³⁺(aq) + 14H₂O.

3CH₃OH(aq) + 3H₂O(aq) → 3HCO₂H(aq) + 12H⁺(aq) + 12e.

So, the net redox reaction will be:

2Cr₂O₇²⁻(aq) + 3CH₃OH(aq) + 16H⁺ → 4Cr³⁺(aq) + 3HCO₂H(aq) + 11H₂O.

Part D: Complete and balance the following equation: BrO₃⁻(aq) + N₂H₄(aq) → Br₂(l) + N₂(g)(acidic solution), Express your answer as a net chemical equation including phases.

To balance and write the net chemical equation, we should write the two-half reactions:

The two half reactions are:

The oxidation reaction: N₂H₄(aq) → N₂(g) + 4e + 4H⁺(aq).

The reduction reaction: 2BrO₃⁻(aq) + 10e + 12H⁺(aq) → Br₂(I) + 6H₂O(aq).

Before adding the reactions number of electron gained must be equal to number of electrons lost. we multiply the oxidation reaction by 5 (5N₂H₄(aq) → 5N₂(g) + 20e + 20H⁺(aq)) and the reduction reaction by 2 (4BrO₃⁻(aq) + 20e + 24H⁺(aq) → 2Br₂(I) + 12H₂O(aq)) to equalize the no. of electrons in the two-half reactions.