Question

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1. Consider the two half-reactions below (attached). The reactant lead, Pb, is
a. oxidized in both
b. reduced in both
c. disproportionated in both
d. oxidized in I and reduced in II

2. In a reduction reaction, the oxidation number becomes more negative due to
a. loss of electrons
b. gain of electrons
c. disproportionation
d. addition of OH- ions

3. The half-reaction 2MnO2 + H2O + 2e- Mn2O3 is missing
a. electrons
b. OH- ions
c. Mn2+ ions
d. O2- ions

The oxidation number of S in SO2−4 is
a. 2
b. +2
c. +4
d. +6

Answer 1

Part 1.

There is no attachment for reaction:

But Option D will be correct because in two half reactions a single compound cannot both oxidized or reduced. Hence it may be oxidized in I and reduced in II .

Part 2.

In a reduction reaction, the oxidation number becomes more negative due to gain of electrons. Option B is correct.

Step-by-step explanation:

The oxidation number of simple ions is equal to the charge on the ion. So more negative means more electrons are added.

Part 3:

The half-reaction 2MnO2 + H2O + 2e- Mn2O3 is missing OH- ions.

Step-by-step explanation:

The full equation is :

2MnO2 + H2O + 2e- → Mn2O3 + 2OH-

So OH- ion were also produced.

Part 4:

The oxidation number of S in SO2 is +4.

For SO2-1 = +3

For SO2-2 = +2

For SO2-4 = +1

Rule 5 says that the sum of oxidation numbers for neutral compounds must be 0.

In charged compounds equal to net charge.