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Iteru [2.4K]
3 years ago
12

For each of the following balanced oxidation-reduction reactions, (i) identify the oxidation numbers for all the elements in the

reactants and products and (ii) state the total number of electrons transferred in each reaction. 1. Identify the oxidation numbers for all the elements in the reactants and products for I2O5(s)+5CO(g)?I2(s)+5CO2(g) 2. Identify the oxidation numbers for all the elements in the reactants and products for 2Hg2+(aq)+N2H4(aq)?2Hg(l)+N2(g)+4H+(aq) 3. Identify the oxidation numbers for all the elements in the reactants and products for 3H2S(aq)+2H+(aq)+2NO?3(aq)?3S(s)+2NO(g)+4H2O(l)
Chemistry
1 answer:
AlladinOne [14]3 years ago
5 0

Answer:

<h3>1. 10 e⁻</h3>

Oxidation numbers

I₂O₅(s): I (5+); O(2-)

CO(g): C(2+); O(2-)

I₂(s): I(0)

CO₂(g): C(4+); O(2-)

<h3>2. 4 e⁻</h3>

Oxidation numbers

Hg²⁺(aq): Hg(2+)

N₂H₄(aq): N(2-); H(1+)

Hg(l): Hg(0)

N₂(g): N(0)

H⁺(aq): H(1+)

<h3>3. 6 e⁻</h3>

Oxidation numbers

H₂S(aq): H(1+); S(2-)

H⁺(aq): H(1+)

NO₃⁻(aq): N(5+); O(2-)

S(s): S(0)

NO(g): N(2+); O(2-)

H₂O(l): H(1+); O(2-)

Explanation:

In order to state the total number of electrons transferred we have to identify both half-reactions for each redox reaction.

1.  I₂O₅(s) + 5 CO(g) → I₂(s) + 5 CO₂(g)

Oxidation: 10 e⁻ + 10 H⁺(aq) + I₂O₅(s) → I₂(s) + 5 H₂O(l)

Reduction: 5 H₂O(l) + 5 CO(g) → 5 CO₂(g) + 10 H⁺(aq) + 10 e⁻

2. 2 Hg²⁺(aq) + N₂H₄(aq) → 2 Hg(l) + N₂(g) + 4 H⁺(aq)

Oxidation: N₂H₄(aq) → N₂(g) + 4 H⁺(aq) + 4 e⁻

Reduction: 2 Hg²⁺(aq) + 4 e⁻ → 2 Hg(l)

3. 3 H₂S(aq) + 2H⁺(aq) + 2 NO₃⁻(aq) → 3 S(s) + 2 NO(g) + 4H₂O(l)

Oxidation: 3 H₂S(aq) → 3 S(s) + 6 H⁺(aq) + 6 e⁻

Reduction: 8 H⁺(aq) + 2 NO₃⁻(aq) + 6 e⁻ → 2 NO(g) + 4 H₂O

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Answer:

  • <u><em>Option C) 1s² 2s² 2p⁵</em></u>

Explanation:

<em>Ground-state</em> is the configuration of the electrons when every electron of the atom occupies the lowest possible energy level.

The rank of energy of the orbitals may be remembered using Aufbau rules, and it is:

1s² < 2s² < 2p⁶ < 3s² < 3p⁶ < 4s² < 3d¹⁰ < 4p⁶ < 5s² < 4d¹⁰ < 5p⁶ < 6s² < 4f¹⁴ < 5d¹⁰ < 6p⁶ < 7s² < 5f¹⁴ < 6d¹⁰ < 7p⁶

Then, to find whether an electron configuration corresponds to a ground-state one, you must check that the previous order is preserved.

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The orbital 1p does not exist. So this one is discarded.

<u>Option B) 1s² 2s⁴ 2p⁶</u>

The maximum number of electrons in an s orbital is 2, so 2s⁴ is impossible, and this option is discarded.

<u>Option C) 1s² 2s² 2p⁵</u>

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This means that 6 moles of H₂O is need for every 1 mole of C₆H₁₂O₆

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