Balancing Redox Reactions: A Step-by-Step Guide

Hey guys! Are you struggling with balancing redox reactions and figuring out those pesky coefficients? Don't worry, you're not alone! This guide will walk you through a specific problem involving copper, nitric acid, nitrogen monoxide, and nitrogen dioxide. We'll break down the steps to determine the coefficients, write the balanced equations, and understand the molar ratios. Let's dive in!

Understanding the Challenge: Copper and Nitric Acid Reactions

Before we jump into the solution, let's understand what we're dealing with. We have two chemical reactions here, both involving redox (reduction-oxidation) processes. Redox reactions are all about the transfer of electrons between chemical species. One species loses electrons (oxidation), and another gains electrons (reduction). Balancing these reactions can seem tricky, but with a systematic approach, it becomes much more manageable.

Our first reaction involves copper (Cu) reacting with nitric acid (HNO3) to produce copper nitrate (Cu(NO3)2), nitrogen monoxide (NO), and water (H₂O). This is a classic example of a metal reacting with an acid, where the metal gets oxidized, and the nitric acid gets reduced. The nitrogen monoxide is released as a gas, indicated by the upward arrow (↑).

The second reaction is simpler: nitrogen monoxide (NO) reacts with oxygen (O2) to form nitrogen dioxide (NO2). This is a straightforward oxidation reaction where nitrogen monoxide gains more oxygen atoms.

The challenge is that these reactions are presented with unknown coefficients (x, y, and z). We need to find the values of these coefficients to balance the equations correctly, ensuring that the number of atoms of each element is the same on both sides of the reaction. We're also given three algebraic equations that relate x, y, and z, which we'll use to solve for these unknowns.

a. Determining the Coefficients x, y, and z

This is where the fun begins! We have a system of equations that we need to solve. Let's recap the equations:

1. x + y + z = 13
2. y - 2x = z
3. x + z = y/2 + 1

Our goal is to find the values of x, y, and z that satisfy all three equations. There are a few ways to solve a system of linear equations, such as substitution, elimination, or using matrices. Let's use the substitution method, as it's quite intuitive for this problem.

Step 1: Simplify the Equations

Before we start substituting, let's simplify the third equation to get rid of the fraction. Multiply both sides of equation (3) by 2:

2(x + z) = y + 2

2x + 2z = y + 2

Now we have:

1. x + y + z = 13
2. y - 2x = z
3. 2x + 2z = y + 2

Step 2: Substitution Time!

Let's use equation (2) to substitute z in equations (1) and (3). This will reduce the number of variables in those equations.

Substitute z = y - 2x into equation (1):

x + y + (y - 2x) = 13

x + y + y - 2x = 13

2y - x = 13 (Equation 4)

Now, substitute z = y - 2x into equation (3):

2x + 2(y - 2x) = y + 2

2x + 2y - 4x = y + 2

-2x + 2y = y + 2

y - 2x = 2 (Equation 5)

Step 3: Solve for x and y

Now we have two equations (4 and 5) with two unknowns (x and y):

1. 2y - x = 13
2. y - 2x = 2

Let's solve this system. We can multiply equation (5) by -2 to eliminate y:

-2(y - 2x) = -2(2)

-2y + 4x = -4

Now add this modified equation to equation (4):

(2y - x) + (-2y + 4x) = 13 + (-4)

3x = 9

x = 3

Great! We found x = 3. Now, substitute this value back into either equation (4) or (5) to find y. Let's use equation (5):

y - 2(3) = 2

y - 6 = 2

y = 8

So, y = 8.

Step 4: Find z

Now that we have x and y, we can easily find z using any of the original equations. Let's use equation (2):

z = y - 2x

z = 8 - 2(3)

z = 8 - 6

z = 2

Solution

We've done it! We found the values of the coefficients:

• x = 3
• y = 8
• z = 2

b. Writing the Balanced Chemical Equations

Now that we have the coefficients, we can write the balanced chemical equations. Let's substitute the values of x, y, and z into the original reactions:

Reaction 1:

2(3)Cu + 2(8)HNO3 → 2(3)Cu(NO3)2 + 4NO↑ + 8H₂O

6Cu + 16HNO3 → 6Cu(NO3)2 + 4NO↑ + 8H₂O

Reaction 2:

2NO + O2 → 2NO2

These are the balanced chemical equations! Notice how the number of atoms of each element is the same on both sides of each equation. This is the essence of a balanced chemical equation – it follows the law of conservation of mass.

c. Indicating the Molar Ratios

The molar ratio is the ratio of the amounts of substances in a chemical reaction. It's derived from the coefficients in the balanced chemical equation. Let's look at the molar ratios for each reaction:

Reaction 1: 6Cu + 16HNO3 → 6Cu(NO3)2 + 4NO↑ + 8H₂O

• Cu : HNO3 Ratio: 6 : 16 (which simplifies to 3:8). This means that for every 3 moles of copper, 8 moles of nitric acid are required.
• Cu : Cu(NO3)2 Ratio: 6 : 6 (which simplifies to 1:1). For every 1 mole of copper that reacts, 1 mole of copper nitrate is produced.
• Cu : NO Ratio: 6 : 4 (which simplifies to 3:2). For every 3 moles of copper that react, 2 moles of nitrogen monoxide are produced.
• Cu : H₂O Ratio: 6 : 8 (which simplifies to 3:4). For every 3 moles of copper that react, 4 moles of water are produced.

Reaction 2: 2NO + O2 → 2NO2

• NO : O2 Ratio: 2 : 1. For every 2 moles of nitrogen monoxide, 1 mole of oxygen is required.
• NO : NO2 Ratio: 2 : 2 (which simplifies to 1:1). For every 1 mole of nitrogen monoxide that reacts, 1 mole of nitrogen dioxide is produced.
• O2 : NO2 Ratio: 1 : 2. For every 1 mole of oxygen that reacts, 2 moles of nitrogen dioxide are produced.

These molar ratios are super important in stoichiometry – the branch of chemistry that deals with the quantitative relationships between reactants and products in chemical reactions. They allow us to predict how much of a product will be formed from a given amount of reactant or how much reactant is needed to produce a certain amount of product.

Conclusion: Mastering Redox Balancing

Balancing redox reactions might seem daunting at first, but by breaking it down into steps and understanding the underlying principles, it becomes much more approachable. We successfully determined the coefficients for the reactions involving copper and nitric acid, wrote the balanced chemical equations, and identified the molar ratios. Remember, practice makes perfect, so keep working on these types of problems, and you'll become a redox balancing pro in no time! Keep up the great work, guys!