Balancing redox equations with oxidation numbers

[gigi9]

Please help me w/ the 2 redox equations. Show me how to assign the oxidation # and how to do it please, especially the 1st problem. The problems are attached. Thank you very much for ur time.

 

[Bystander]

Are the subscripts EXACTLY what you were given in the problem statement? (I know they don't agree with the original problem statement, but you may be dealing with an instructional transcription error.)

The first reaction should show ferric hydroxide (no net charge) reacting with chrome-three (Cr(III), sometimes chromous) hydroxide to produce ferrous hydroxide and chromate ion. Get you started? You have played the "game" of adding hydroxide ions (or hydrogen ion) to one side or the other for basic (or acidic) conditions?

 

[M98Ranger]

Hi there,

I would be happy to assist you with balancing redox equations. Before we start, it's important to understand the concept of oxidation numbers. Oxidation numbers are assigned to each element in a compound or ion to indicate its degree of oxidation or reduction. In general, oxidation numbers follow these rules:

1. The oxidation number of an element in its elemental form is always 0.
2. The sum of oxidation numbers in a neutral compound is always 0.
3. The sum of oxidation numbers in a polyatomic ion is equal to the charge of the ion.

Now, let's look at the first problem attached. It involves the reaction between potassium dichromate (K2Cr2O7) and iron(II) sulfate (FeSO4) in an acidic solution to form chromium(III) sulfate (Cr2(SO4)3) and iron(III) sulfate (Fe2(SO4)3).

Step 1: Assign oxidation numbers to each element in the reaction.

K2Cr2O7: K has an oxidation number of +1, Cr has an oxidation number of +6, and O has an oxidation number of -2.
FeSO4: Fe has an oxidation number of +2, S has an oxidation number of +6, and O has an oxidation number of -2.
Cr2(SO4)3: Cr has an oxidation number of +3, S has an oxidation number of +6, and O has an oxidation number of -2.
Fe2(SO4)3: Fe has an oxidation number of +3, S has an oxidation number of +6, and O has an oxidation number of -2.

Step 2: Determine which elements are being oxidized and reduced.

In this reaction, iron is being oxidized from an oxidation number of +2 to +3, while chromium is being reduced from an oxidation number of +6 to +3.

Step 3: Write out the half-reactions.

Oxidation half-reaction:
Fe → Fe3+ + e-

Reduction half-reaction:
Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O

Step 4: Balance the number of electrons in each half-reaction.

Multiply the oxidation half-reaction by 6 and the reduction half-reaction by 3 to balance the number of electrons.

6Fe →