Here, we’ll use the half-reactions we came up with and balanced in Part 1 to build an equation for...

Redox Equations in Basic Solution

Half Reaction Method

Part 2The Redox Equation

Here, we’ll use the half-reactions we came up with and balanced in Part 1 to build an equation for the overall Redox Reaction taking place. We’ll balance it in acid solution here. Then in Part 3, we will convert it to Basic Solution

Here are the two balanced half-reactions we obtained in Part 1.

Write a balanced redox equation in basic solution, given the skeleton equation:

22 3 2NH OH HPO N P

32NH OH N 2H O 2H 2e

HPO 5H 3e P 3H 2O

22 3 2 2 26NH OH 2HPO 10H 3N 6H O 6H 2P 6H O

22 3 2 2

6NH OH 2HPO 4H 3N 12H O 2P

4H O 4H 4OH

6NH OH 2HPO 3N 8H O 2P 4OH

The 2 half-reactions webalanced in Part 1

In order to add these up and obtain the overall redox equation, we must look at the electrons.

22 3 2NH OH HPO N P

2NH OH N 2H O 2H 3

HPO 5H

3e P 3H O 2

22 3 2 2

4H O 4H 4OH

We see that 3 electrons and gained by the lower half-reaction, and (click) 2 electrons are lost by the top half-reaction.

22 3 2NH OH HPO N P

2NH OH N 2H O 2H 3

HPO 5H

3e P 3H O 2

22 3 2 2

4H O 4H 4OH

So the electrons gained are NOT equal to the electrons lost. We must multiply these half-reactions by factors which will make the electrons gained equal to those lost

22 3 2NH OH HPO N P

2NH OH N 2H O 2H 3

HPO 5H

3e P 3H O 2

22 3 2 2

4H O 4H 4OH

e–’s gained ≠ e–’s gained

The lowest common multiple of 3 and 2, is 6. Therefore we multiply by factors which will make electrons gained and electrons lost, both equal to 6.

22 3 2NH OH HPO N P

2NH OH N 2H O 2H 3

HPO 5H

3e P 3H O 2

22 3 2 2

4H O 4H 4OH

Lowest common multiple of 3 and 2 is 6

So we multiply the top half-reaction by 3,

22 3 2NH OH HPO N P

32NH OH N 2H O 2H

HPO 5H 3e P 3H

22 3 2 2

4H O 4H 4OH

And the bottom half-reaction by 2

22 3 2NH OH HPO N P

32NH OH N 2H O 2H

HPO 5H P 33 H

22 3 2 2

4H O 4H 4OH

So 2 × 3 = 6 electrons are gained by the bottom half-reaction

22 3 2NH OH HPO N P

32NH OH N 2H O 2H

HPO 5H P 33 H

22 3 2 2

4H O 4H 4OH

6 e– gaine

And 3 × 2 = 6 electrons are lost by the top half-reaction.

22 3 2NH OH HPO N P

32NH OH N 2H O 2H

HPO 5H P 33 H

22 3 2 2

4H O 4H 4OH

6 e– lost

So, because electrons gained are now equal to electrons lost,

22 3 2NH OH HPO N P

32NH OH N 2H O 2H

HPO 5H P 33 H

22 3 2 2

4H O 4H 4OH

6 e– lost

6 e– gaine

The electrons can be cancelled from the half-reactions

22 3 2NH OH HPO N P

32NH OH N 2H O 2H

HPO 5H P 33 H

22 3 2 2

4H O 4H 4OH

So now we’re left with these.

22 3 2NH OH HPO N P

32NH OH N 2H O 2H

HPO 5H P 33 H

22 3 2 2

4H O 4H 4OH

At this point, we add what we have on the left and right of the arrows to obtain the equation for the overall redox reaction.

22 3 2NH OH HPO N P

32NH OH N 2H O 2H

HPO 5H P 33 H

22 3 2 2

4H O 4H 4OH

Starting on the top left, we have (click) 3 × 2 NH2OH

22 3 2NH OH HPO N P

32NH O N 2H O 2H

HPO 5H P 3H O

22 3 2 2

4H O 4H 4OH

Which gives us 6 NH2OH

22 3 2NH OH HPO N P

32NH O N 2H O 2H

HPO 5H P 3H O

23 22 2 22HPO 10H 3N 6H O 66N H 2 OH OH P 6H 2

2 3 2 2

22 3 2 2

4H O 4H 4OH

And on the bottom left, we have 2 × HPO3 (2 minus)

22 3 2NH OH HPO N P

2 32NH OH N 2H O 2H

5H P 3H

HPO e O3 2

23 22 2 22HPO 10H 3N 6H O 66N H 2 OH OH P 6H 2

2 3 2 2

22 3 2 2

4H O 4H 4OH

Which we’ll add here,

22 3 2NH OH HPO N P

2 32NH OH N 2H O 2H

5H P 3H

HPO e O3 2

2 2 2 2236NH OH 10H 3N 6H O 6H2HP 2P 6HO O 2

2 3 2 2

22 3 2 2

4H O 4H 4OH

And 2 × 5 H+,

22 3 2NH OH HPO N P

32NH OH N 2H O 2H

HPO P 3H O

H 3 2e5

2 2 2 2236NH OH 10H 3N 6H O 6H2HP 2P 6HO O 2

2 3 2 2

22 3 2 2

4H O 4H 4OH

Which equals 10 H+.

22 3 2NH OH HPO N P

32NH OH N 2H O 2H

HPO P 3H O

H 3 2e5

22 3 2 2 26NH OH 2HPO 3N 6H O 6H 2P1 6H O0H

22 3 2 2

4H O 4H 4OH

Now for the right side. On the top right we have (click) 3 × 1 N2,

22 3 2NH OH HPO N P

32NH OH 2H O 2H

HPO 5H P 3

3 Oe 2

2 3 2 222 3N6NH OH 2HPO 10H 6H O 6H 2P 6H O 2

2 3 2 2

22 3 2 2

4H O 4H 4OH

Which is 3 N2,

22 3 2NH OH HPO N P

32NH OH 2H O 2H

HPO 5H P 3

3 Oe 2

2 3 2 2

22 3 2 2

4H O 4H 4OH

3 × 2 H2O,

22 3 2NH OH HPO N P

2 32NH OH N 2H

HPO 5H P 3H O

2 3 2 2

22 3 2 2

4H O 4H 4OH

Which is equal to 6 H2O,

22 3 2NH OH HPO N P

2 32NH OH N 2H

HPO 5H P 3H O

22 3 2 226NH OH 2HPO 10H 3N 6H6H O 2P 6H O

22 3 2 2

4H O 4H 4OH

And 3 times 2 H+,

22 3 2NH OH HPO N P

32NH OH N 2H O

HPO 5H P 3

22 3 2 2 26NH OH 2HPO 10H 3 6HN 6H O 2P 6H O

22 3 2 2

4H O 4H 4OH

Which equals 6 H+.

22 3 2NH OH HPO N P

32NH OH N 2H O

HPO 5H P 3

22 3 2 2 26NH OH 2HPO 10H 3 6HN 6H O 2P 6H O

22 3 2 2

4H O 4H 4OH

On the bottom right, we have 2 × P

22 3 2NH OH HPO N P

32NH OH N 2H O 2H

HPO 5H 3P H

3 Oe 2

22 3 2 2 26NH OH 2HPO 10H 3N 6H O 6H 2 HP 6 O

22 3 2 2

4H O 4H 4OH

Which is equal to 2 P

22 3 2NH OH HPO N P

32NH OH N 2H O 2H

HPO 5H 3P H

3 Oe 2

22 3 2 2 26NH OH 2HPO 10H 3N 6H O 6H 2 HP 6 O

22 3 2 2

4H O 4H 4OH

And 2 × 3 H2O,

22 3 2NH OH HPO N P

32NH O

H N 2H O 2H

HPO 5H P H 2e O

2 3 2 26NH OH 2HPO 10H 3N 6H O 6H 2 HP 6 O 2

2 3 2 2

22 3 2 2

4H O 4H 4OH

Which is equal to 6 H2O.

22 3 2NH OH HPO N P

32NH O

H N 2H O 2H

HPO 5H P H 2e O

2 3 2 26NH OH 2HPO 10H 3N 6H O 6H 2 HP 6 O 2

2 3 2 2

22 3 2 2

4H O 4H 4OH

Now we have the balanced redox equation.

22 3 2NH OH HPO N P

32NH O

H N 2H O 2H

HPO 5H P H 2e O

22 3 2 2

4H O 4H 4OH

Balanced Redox Equation

But it has H+ ions on both sides,

22 3 2NH OH HPO N P

32NH O

H N 2H O 2H

HPO 5H P H 2e O

22 3 2 2

4H O 4H 4OH

And two sets of water molecules on the right side,

22 3 2NH OH HPO N P

32NH O

H N 2H O 2H

HPO 5H P H 2e O

22 3 2 2

4H O 4H 4OH

So this equation must be simplified

22 3 2NH OH HPO N P

32NH O

H N 2H O 2H

HPO 5H P H 2e O

22 3 2 2

4H O 4H 4OH

Must Be Simplified

We start by removing 6 H+ ions from both sides,

22 3 2NH OH HPO N P

32NH O

H N 2H O 2H

HPO 5H P H 2e O

22 3 2 2 26NH OH 2HPO 3N 6H O 2P 6H O10H 6H

22 3 2 2

4H O 4H 4OH

Remove 6 H+ ions from Both Sides

Removing 6 H+ ions from the right side, leaves us with none

22 3 2NH OH HPO N P

32NH O

H N 2H O 2H

HPO 5H P H 2e O

22 3 2 2

4H O 4H 4OH

And Removing 6 H+ ions from the left side leaves us with 10 – 6 (click), or 4 H+ ion

22 3 2NH OH HPO N P

32NH O

H N 2H O 2H

HPO 5H P H 2e O

22 3 2 2

4H O 4H 4OH

We simplify the water by adding the two sets of 6 H2O molecules on the right side

22 3 2NH OH HPO N P

32NH O

H N 2H O 2H

HPO 5H P H 2e O

2 3 26NH OH 2HPO 10H 3N 6H6H O O2P 6H 2

2 3 2 2

22 3 2 2

4H O 4H 4OH

Add up the two sets of H2O molecules

To give us 12 water molecules altogether.

22 3 2NH OH HPO N P

32NH O

H N 2H O 2H

HPO 5H P H 2e O

2 3 26NH OH 2HPO 10H 3N 6H6H O O2P 6H 2

2 3 2 2

22 3 2 2

4H O 4H 4OH

6H2O + 6H2O = 12H2O

We’ll re-write this equation in more compact form here

22 3 2NH OH HPO N P

32NH O

H N 2H O 2H

HPO 5H P H 2e O

22 3 2

4H O 4H 4OH

6NH OH

6NH OH 2HPO 4H 3N 12H O

2HPO 3N 8H O

2P 4OH

So now we have the redox equation balanced in acid solution. At this point, pause the video and check to see that all atoms are balanced and total ionic charge is balanced

22 3 2NH OH HPO N P

32NH O

H N 2H O 2H

HPO 5H P H 2e O

22 3 2

4H O 4H 4OH

6NH OH

2HPO 3N 8H O

2P 4OH

Balanced Redox Equation in Acid Solution

However, the original question wants us to balance this equation in BASIC solution

22 3 2NH OH HPO N P

32NH O

H N 2H O 2H

HPO 5H P H 2e O

22 3 2

4H O 4H 4OH

6NH OH

2HPO 3N 8H O

2P 4OH

Balanced Redox Equation in Acid Solution

So this equation needs to be converted to Basic Solution

22 3 2NH OH HPO N P

32NH O

H N 2H O 2H

HPO 5H P H 2e O

22 3 2

4H O 4H 4OH

6NH OH

2HPO 3N 8H O

2P 4OH

Convert to Basic Solution

In the next video, which is Part 3, we will convert this redox equation to basic solution.

22 3 2NH OH HPO N P

32NH O

H N 2H O 2H

HPO 5H P H 2e O

22 3 2

4H O 4H 4OH

6NH OH

2HPO 3N 8H O

2P 4OH

In PART 3 (the next video) we’ll Convert this to Basic

Solution

Here, we’ll use the half-reactions we came up with and balanced in Part 1 to build an equation for...

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