Experiment 1 Pre Lab 1. During this experiment, solid or aqueous forms of the so

Question

Experiment 1 Pre Lab 1. During this experiment, solid or aqueous forms of the sodium salts shown below will be reacted individually with the reagents H2So. AgNO, and BaCl2. In this way, the reactivity of each anion will be determined. For those data cells not marked out in the table below, predict products and what should be observed experimentally (gas, precipitate, etc). NOTE: Most SO, salts are except those of Na ,K and NH., while most Noa salts are solubleexcept AgNO which is sparingly soluble. Sodium Salt of Anion Dilute Haso. (aq) BaCl2(aq) AgNO, Precipitate Formed? Precipitate Formed? Gas Evolved? NaNO, Predicted products? NaNO, Predicted products? Namco Predicted products? Naso, Predicted products? Na,SO, Predicted products? NasPO. Predicted products? Na,Cro, Predicted products? Experiment 1 I Identification ef Anions

Explanation / Answer

2.

a.

An oxidizing agent is a compound which oxidizes other compounds, and in this process it itself gets reduced.

In terms of electron transfer, oxidation is termed as loss of electrons while reduction is termed as gain of electrons.

Thus an oxidizing agent accepts electrons from other compound, which means that the other compound undergoes oxidation while the oxidizing agent undergoes reduction.

b.

Oxidation number of N in NO2- : +3

Calculation method:

(oxidation number of N * no. of atoms of N + oxidation number of O * no. of atoms of O) = Net charge on molecule

Oxidation no. of O is almost always -2.

Thus, x*1 + (-2)*2 = -1

Thus, x = +3

Similarly,

Oxidation number of N in NO3- : +5

Oxidation number of C in CO32- : +4

Oxidation number of S in SO32- : +4

Oxidation number of S in SO42- : +6

Oxidation number of P in PO43- : +5

Oxidation number of Cr in CrO42- : +6

c.

In part (b) above we see that in SO32- ion, S has an oxidation state of +4, while it can have a higher oxidation state of +6 as well.

Thus it can be oxidized by MnO4- by the following reaction:
MnO4- + 4SO32- ---> 4SO42- + Mn2+

d.

Only BrO4- cannot be oxidized while all others can be , because in BrO4- Br is already having its highest oxidation state of +7, while it is having +1, +3 and +5 states in other compounds which can thus be oxidized.

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