Balancing Redox Chemical Reactions by the Half-Reaction Method Assign oxidation

Question

Balancing Redox Chemical Reactions by the Half-Reaction Method Assign oxidation numbers to each atom. 2. Splt the skeleton equation into two half-reactions, proceeding as follows, Note the species containing the elemer proceeding as follows. Note the species con that increases in oxidation number and write those species to give the oxidation half-reaction. Similarly, note t species containing the element that decreases in oxidation number, and write the reduction half-reaction. Complete and balance each half-reaction a. Balance all atoms except O and H. b. Balance O atoms by adding H:O's to one side of the equation. c. Balance H atoms by adding H" ions to one side of the equation. d. Balance electric charge by adding electrons (e') to the more positive side. Combine the two half-reactions to obtain the final balanced oxidation- reduction equation. a. Multiply each half-reaction by a factor such that when the half-reactions are added, the electrons cancel. (Electrons cannot appear in the final equation.) b. Simplify the balanced equation by canceling species that occur on both sides, and reduce the coefficients to smallest whole numbers. Check that the equation is indeed balanced. 3. 4. Example: Zn(s)+ NOj (a)n2 (aa) +NH (aq)

Explanation / Answer

Answer:

The sum of the oxidation numbers in a neutral compound is zero. 10. The sum of the oxidation numbers in a polyatomic ion is equal to the charge on the ion. Theoxidation number of the sulfur atom in the SO42- ion must be +6, for example, because the sum of the oxidation numbers of the atoms in this ion must equal -2

Acidic Conditions

Acidic conditions usually implies a solution with an excess of H+ concentration, hence making the solution acidic. The balancing starts by separating the reaction into half-reactions.

However, instead of immediately balancing the electrons, balance all the elements in the half-reactions that are not hydrogen and oxygen. Then, add H2O molecules to balance any oxygen atoms. Next, balance the hydrogen atoms by adding protons (H+). Now, balance the charge by adding electrons and scale the electrons (multiply by the lowest common multiple) so that they will cancel out when added together. Finally, add the two half-reactions and cancel out common terms.

BALANCING IN A ACID SOLUTION

Balance the following redox reaction in acidic conditions.

Cr2O2?7(aq)+HNO2(aq)?Cr3+(aq)+NO?3(aq)(1.13)(1.13)Cr2O72?(aq)+HNO2(aq)?Cr3+(aq)+NO3?(aq)

Solution

Step 1: Separate the half-reactions. The table provided does not have acidic or basic half-reactions, so just write out what is known.

Cr2O2?7(aq)?Cr3+(aq)(1.14)(1.14)Cr2O72?(aq)?Cr3+(aq)

HNO2(aq)?NO?3(aq)(1.15)(1.15)HNO2(aq)?NO3?(aq)

Step 2: Balance elements other than O and H. In this example, only chromium needs to be balanced. This gives:

Cr2O2?7(aq)?2Cr3+(aq)(1.16)(1.16)Cr2O72?(aq)?2Cr3+(aq)

HNO2(aq)?NO?3(aq)(1.17)(1.17)HNO2(aq)?NO3?(aq)

Step 3: Add H2O to balance oxygen. The chromium reaction needs to be balanced by adding 7 H2O molecules. The other reaction also needs to be balanced by adding one water molecule. This yields:

Cr2O2?7(aq)?2Cr3+(aq)+7H2O(l)(1.18)(1.18)Cr2O72?(aq)?2Cr3+(aq)+7H2O(l)

HNO2(aq)+H2O(l)?NO?3(aq)(1.19)(1.19)HNO2(aq)+H2O(l)?NO3?(aq)

Step 4: Balance hydrogen by adding protons (H+). 14 protons need to be added to the left side of the chromium reaction to balance the 14 (2 per water molecule * 7 water molecules) hydrogens. 3 protons need to be added to the right side of the other reaction.

14H+(aq)+Cr2O2?7(aq)?2Cr3+(aq)+7H2O(l)(1.20)(1.20)14H+(aq)+Cr2O72?(aq)?2Cr3+(aq)+7H2O(l)

HNO2(aq)+H2O(l)?3H+(aq)+NO?3(aq)(1.21)(1.21)HNO2(aq)+H2O(l)?3H+(aq)+NO3?(aq)

Step 5: Balance the charge of each equation with electrons. The chromium reaction has (14+) + (2-) = 12+ on the left side and (2 * 3+) = 6+ on the right side. To balance, add 6 electrons (each with a charge of -1) to the left side:

6e?+14H+(aq)+Cr2O2?7(aq)?2Cr3+(aq)+7H2O(l)(1.22)(1.22)6e?+14H+(aq)+Cr2O72?(aq)?2Cr3+(aq)+7H2O(l)

For the other reaction, there is no charge on the left and a (3+) + (-1) = 2+ charge on the right. So add 2 electrons to the right side:

HNO2(aq)+H2O(l)?3H+(aq)+NO?3(aq)+2e?(1.23)(1.23)HNO2(aq)+H2O(l)?3H+(aq)+NO3?(aq)+2e?

Step 6: Scale the reactions so that the electrons are equal. The chromium reaction has 6e- and the other reaction has 2e-, so it should be multiplied by 3. This gives:

3?[HNO2(aq)+H2O(l)?3H+(aq)+NO?3(aq)+2e?]?(1.24)(1.24)3?[HNO2(aq)+H2O(l)?3H+(aq)+NO3?(aq)+2e?]?

3HNO2(aq)+3H2O(l)?9H+(aq)+3NO?3(aq)+6e?(1.25)(1.25)3HNO2(aq)+3H2O(l)?9H+(aq)+3NO3?(aq)+6e?

6e?+14H+(aq)+Cr2O2?7(aq)?2Cr3+(aq)+7H2O(l).(1.26)(1.26)6e?+14H+(aq)+Cr2O72?(aq)?2Cr3+(aq)+7H2O(l).

Step 7: Add the reactions and cancel out common terms.

[3HNO2(aq)+3H2O(l)?9H+(aq)+3NO?3(aq)+6e?]+(1.27)(1.27)[3HNO2(aq)+3H2O(l)?9H+(aq)+3NO3?(aq)+6e?]+

[6e?+14H+(aq)+Cr2O2?7(aq)?2Cr3+(aq)+7H2O(l)]=(1.28)(1.28)[6e?+14H+(aq)+Cr2O72?(aq)?2Cr3+(aq)+7H2O(l)]=

3HNO2(aq)+3H2O(l)+6e?+14H+(aq)+Cr2O2?7(aq)?9H+(aq)+3NO?3(aq)+6e?+2Cr3+(aq)+7H2O(l)

3HNO2(aq)+3H2O(l)+6e?+14H+(aq)+Cr2O72?(aq)?9H+(aq)+3NO3?(aq)+6e?+2Cr3+(aq)+7H2O(l)

The electrons cancel out as well as 3 water molecules and 9 protons. This leaves the balanced net reaction of:

3HNO2(aq)+5H+(aq)+Cr2O2?7(aq)?3NO?3(aq)+2Cr3+(aq)+4H2O(l)

Thank You..!

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