Redox Reaction

[Lilac12]

"Write half-equations and hence a balance overall ionic equation for the reaction of acidified potassium dichromate solution and a solution of iron (ll) nitrate. (Note the potassium and nitrate ions are spectator ions)."

I'd really appreciate a step-by-step explanation of how to get to the final answer!

[ChenBeier]

1. Build redox pairs Cr₂O₇ 2-/ Cr 3+ and Fe 2+/ Fe 3+
2. Check the condition is it acidic, neutral or alcaline
In case of acidic we add H+( H₃O+) to the oxidiser to get the oxygen away, water is developed
In case of neutral or alcaline add water to the oxidiser and OH- is developed
If the reducer also contain oxygen then the same thing vise versa means add water get H+ or add OH- and water is developed.
3. Develop the half reaction

Cr₂O₇ 2- + 14 H+=> 2 Cr 3+ + 7 H₂O
And Fe 2+=> Fe 3+
4. Balance the charges with electrons e-
Cr₂O₇ 2- + 14 H+ + 6 e-=> 2 Cr 3+ + 7 H₂O
Fe 2+ => Fe 3+ + e-

5. Build the least common multiple of the electrons
In this case its 6
Multiply the equation
Cr₂O₇ 2- + 14 H+ + 6 e-=> 2 Cr 3+ + 7 H₂O
6 Fe 2+ => 6 Fe 3+ + 6 e-

6. Add both and eliminate the electrons, H+ and water if necessary?

Cr₂O₇ 2- + 14 H+ + 6 Fe 2+ => 2 Cr 3+ + 6 Fe 3+ +7 H₂O

Finished

[Lilac12]

Thank you so much for replying it was really helpful!

I'm not sure if this is a stupid question, but the part that I was struggling with was the very first step that you did. How did you get from Iron 2+ to Iron 3+ and the same with the Cr turning into Cr 3+ ?

[ChenBeier]

It's knowledge to learn. Read the chemical behavior of iron ,chromium and it's compounds, the same for manganese.

[Lilac12]

Thank you, I think I understand how you got to the final answer now. I really appreciate you taking time out of your day to write out an explanation.

[Violetica]

1. Build redox pairs Cr₂O₇ 2-/ Cr 3+ and Fe 2+/ Fe 3+
2. Check the condition is it acidic, neutral or alcaline
In case of acidic we add H+( H₃O+) to the oxidiser to get the oxygen away, water is developed
In case of neutral or alcaline add water to the oxidiser and OH- is developed
If the reducer also contain oxygen then the same thing vise versa means add water get H+ or add OH- and water is developed.
3. Develop the half reaction

Cr₂O₇ 2- + 14 H+=> 2 Cr 3+ + 7 H₂O
And Fe 2+=> Fe 3+
4. Balance the charges with electrons e-
Cr₂O₇ 2- + 14 H+ + 6 e-=> 2 Cr 3+ + 7 H₂O
Fe 2+ => Fe 3+ + e-

5. Build the least common multiple of the electrons
In this case its 6
Multiply the equation
Cr₂O₇ 2- + 14 H+ + 6 e-=> 2 Cr 3+ + 7 H₂O
6 Fe 2+ => 6 Fe 3+ + 6 e-

6. Add both and eliminate the electrons, H+ and water if necessary?

Cr₂O₇ 2- + 14 H+ + 6 Fe 2+ => 2 Cr 3+ + 6 Fe 3+ +7 H₂O

Finished

[Violetica]

Excellent explanation

[sportav]

1. Build redox pairs Cr₂O₇ 2-/ Cr 3+ and Fe 2+/ Fe 3+
2. Check the condition is it acidic, neutral or alcaline
In case of acidic we add H+( H₃O+) to the oxidiser to get the oxygen away, water is developed
In case of neutral or alcaline add water to the oxidiser and OH- is developed
If the reducer also contain oxygen then the same thing vise versa means add water get H+ or add OH- and water is developed.
3. Develop the half reaction

Cr₂O₇ 2- + 14 H+=> 2 Cr 3+ + 7 H₂O
And Fe 2+=> Fe 3+
4. Balance the charges with electrons e-
Cr₂O₇ 2- + 14 H+ + 6 e-=> 2 Cr 3+ + 7 H₂O
Fe 2+ => Fe 3+ + e-

5. Build the least common multiple of the electrons
In this case its 6
Multiply the equation
Cr₂O₇ 2- + 14 H+ + 6 e-=> 2 Cr 3+ + 7 H₂O
6 Fe 2+ => 6 Fe 3+ + 6 e-

6. Add both and eliminate the electrons, H+ and water if necessary?

Cr₂O₇ 2- + 14 H+ + 6 Fe 2+ => 2 Cr 3+ + 6 Fe 3+ +7 H₂O

Finished

WOWWW, that's sound great