Ireally need someone to me with just answer what you can as long as you are i will mark brainiliest to those who answer the most. 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

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 .

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

Explanation:

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

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

Explanation:

The full equation is :

                        2MnO2 + H2O + 2e- → Mn2O3  +  2OH-

So OH- ion were also produced.

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.

answer: they produce more slow than unicellular organisms.