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How do you balance a chemical equation in a basic medium if there are 3 reactants and 4 products? How do I know which to use for the half reactions? Example: CrI3 + KOH + Cl2 —> K2Cr2O7 + KIO4 + KCl +H2O
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Lore Murdock
How do you balance a chemical equation in a basic medium if there are 3 reactants and 4 products? How do I know which to use for the half reactions? Example: CrI3 + KOH + Cl2 —> K2Cr2O7 + KIO4 + KCl +H2O
I will show two different methods to solve this.
First is an algebraic method to balance this without using redox half-reactions. Assign letter coefficients to each reactant and product a CrI3 + b KOH + c Cl2 => d K2Cr2O7 + e KIO4 + f KCl + g H2O
Set up equations based on coefficients and atom balance Balance Cr atoms: a = 2d Balance I atoms: 3a = e Balance K atoms: b = 2d + e + f Balance O atoms: b = 7d + 4e + g Balance H atoms: b = 2g Balance Cl atoms: 2c = f
The variables b, d, and e appear in 3 of the equations, so pick one of them (I’ll pick d) and set it equal to 1. So d = 1. Now let’s substitute and try to solve for the others. a = 2d, so a = 2 3a = e, so e = 6
b = 2d + e + f, so b = 2 + 6 + f or b = 8 + f and b = 7d + 4e + g, so b = 7 + 24 + g or b = 31 + g b = 2g, so 2g = 31 + g and g = 31 b = 2g, so b = 62 b = 8 + f, so 62 = 8 + f, and f = 54
2c = f, so 2c = 54, and c = 27 b = 2d + e + f, so 62 = 2d + 6 + 54, 2d = 2, and d = 1
Now put the coefficients into the equation 2 CrI3 + 62 KOH + 27 Cl2 => K2Cr2O7 + 6 KIO4 + 54 KCl + 31 H2O
First is an algebraic method to balance this without using redox half-reactions. Assign letter coefficients to each reactant and product a CrI3 + b KOH + c Cl2 => d K2Cr2O7 + e KIO4 + f KCl + g H2O
Set up equations based on coefficients and atom balance Balance Cr atoms: a = 2d Balance I atoms: 3a = e Balance K atoms: b = 2d + e + f Balance O atoms: b = 7d + 4e + g Balance H atoms: b = 2g Balance Cl atoms: 2c = f
The variables b, d, and e appear in 3 of the equations, so pick one of them (I’ll pick d) and set it equal to 1. So d = 1. Now let’s substitute and try to solve for the others. a = 2d, so a = 2 3a = e, so e = 6
b = 2d + e + f, so b = 2 + 6 + f or b = 8 + f and b = 7d + 4e + g, so b = 7 + 24 + g or b = 31 + g b = 2g, so 2g = 31 + g and g = 31 b = 2g, so b = 62 b = 8 + f, so 62 = 8 + f, and f = 54
2c = f, so 2c = 54, and c = 27 b = 2d + e + f, so 62 = 2d + 6 + 54, 2d = 2, and d = 1
Now put the coefficients into the equation 2 CrI3 + 62 KOH + 27 Cl2 => K2Cr2O7 + 6 KIO4 + 54 KCl + 31 H2O
I will show two different methods to solve this.
First is an algebraic method to balance this without using redox half-reactions.
Assign letter coefficients to each reactant and product
a CrI3 + b KOH + c Cl2 => d K2Cr2O7 + e KIO4 + f KCl + g H2O
Set up equations based on coefficients and atom balance
Balance Cr atoms:
a = 2d
Balance I atoms:
3a = e
Balance K atoms:
b = 2d + e + f
Balance O atoms:
b = 7d + 4e + g
Balance H atoms:
b = 2g
Balance Cl atoms:
2c = f
The variables b, d, and e appear in 3 of the equations, so pick one of them (I’ll pick d) and set it equal to 1. So d = 1. Now let’s substitute and try to solve for the others.
a = 2d, so a = 2
3a = e, so e = 6
b = 2d + e + f, so b = 2 + 6 + f or b = 8 + f and
b = 7d + 4e + g, so b = 7 + 24 + g or b = 31 + g
b = 2g, so 2g = 31 + g and g = 31
b = 2g, so b = 62
b = 8 + f, so 62 = 8 + f, and f = 54
2c = f, so 2c = 54, and c = 27
b = 2d + e + f, so 62 = 2d + 6 + 54, 2d = 2, and d = 1
Now put the coefficients into the equation
2 CrI3 + 62 KOH + 27 Cl2 => K2Cr2O7 + 6 KIO4 + 54 KCl + 31 H2O
Check
atoms: Cr(2), I(6), K(62), O(62), H(62), Cl(54)
electrical charge: 0
Second is the redox half-reaction method:
1. Remove K+ and OH- from both sides of equation (removing OH- from H2O leaves H+)
CrI3 + Cl2 => Cr2O7(2-) + IO4(-) + Cl- + H+
2. Both Cr(+3 -> +6) and I (-1 -> +7) are oxidized, but fortunately they end up in two separate products, so the oxidation half reaction is
CrI3 => Cr2O7(2-) + IO4(-) + H+
2 CrI3 => Cr2O7(2-) + 6 IO4(-) + H+ (balance atoms whose oxidation number change)
31 H2O + 2 CrI3 => Cr2O7(2-) + 6 IO4(-) + H+ (balance O with H2O)
31 H2O + 2 CrI3 => Cr2O7(2-) + 6 IO4(-) + 62 H+ (balance H with H+)
31 H2O + 2 CrI3 => Cr2O7(2-) + 6 IO4(-) + 62 H+ + 54 e- (balance charge with electrons)
3. The reduction half-reaction is
Cl2 => Cl-
Cl2 => 2 Cl- (balance atoms whose oxidation number change)
2e- + Cl2 => 2 Cl- (balance charge with electrons)
27(2e- + Cl2 => 2 Cl-) (make electron gain equal electron loss)
54 e- + 27 Cl2 => 54 Cl-
4. Combine the two half-reactions
27 Cl2 + 31 H2O + 2 CrI3 => Cr2O7(2-) + 6 IO4(-) + 62 H+ + 54 Cl-
5. Add 62 OH- to each side (to neutralize H+ and make a basic solution)
62 OH- + 27 Cl2 + 31 H2O + 2 CrI3 => Cr2O7(2-) + 6 IO4(-) + 62 H+ + 62 OH- + 54 Cl-
62 OH- + 27 Cl2 + 31 H2O + 2 CrI3 => Cr2O7(2-) + 6 IO4(-) + 62 H2O + 54 Cl- (H+ + OH- => H2O)
62 OH- + 27 Cl2 + 2 CrI3 => Cr2O7(2-) + 6 IO4(-) + 31 H2O + 54 Cl- (simplify)
6. Add 62 K+ to each side (to recreate KOH, K2Cr2O7, KIO4, and KCl)
62 K+ + 62 OH- + 27 Cl2 + 2 CrI3 => 2 K+ + Cr2O7(2-) + 6 K+ + 6 IO4(-) + 31 H2O + 54 K+ + 54 Cl-
62 KOH + 27 Cl2 + 2 CrI3 => K2Cr2O7 + 6 KIO4 + 31 H2O + 54 KCl (combine ions)
Same as above. I think the half-reaction method is easier in this case, but you can decide…
I will show two different methods to solve this.
First is an algebraic method to balance this without using redox half-reactions.
Assign letter coefficients to each reactant and product
a CrI3 + b KOH + c Cl2 => d K2Cr2O7 + e KIO4 + f KCl + g H2O
Set up equations based on coefficients and atom balance
Balance Cr atoms:
a = 2d
Balance I atoms:
3a = e
Balance K atoms:
b = 2d + e + f
Balance O atoms:
b = 7d + 4e + g
Balance H atoms:
b = 2g
Balance Cl atoms:
2c = f
The variables b, d, and e appear in 3 of the equations, so pick one of them (I’ll pick d) and set it equal to 1. So d = 1. Now let’s substitute and try to solve for the others.
a = 2d, so a = 2
3a = e, so e = 6
b = 2d + e + f, so b = 2 + 6 + f or b = 8 + f and
b = 7d + 4e + g, so b = 7 + 24 + g or b = 31 + g
b = 2g, so 2g = 31 + g and g = 31
b = 2g, so b = 62
b = 8 + f, so 62 = 8 + f, and f = 54
2c = f, so 2c = 54, and c = 27
b = 2d + e + f, so 62 = 2d + 6 + 54, 2d = 2, and d = 1
Now put the coefficients into the equation
2 CrI3 + 62 KOH + 27 Cl2 => K2Cr2O7 + 6 KIO4 + 54 KCl + 31 H2O
Check
atoms: Cr(2), I(6), K(62), O(62), H(62), Cl(54)
electrical charge: 0
Second is the redox half-reaction method:
1. Remove K+ and OH- from both sides of equation (removing OH- from H2O leaves H+)
CrI3 + Cl2 => Cr2O7(2-) + IO4(-) + Cl- + H+
2. Both Cr(+3 -> +6) and I (-1 -> +7) are oxidized, but fortunately they end up in two separate products, so the oxidation half reaction is
CrI3 => Cr2O7(2-) + IO4(-) + H+
2 CrI3 => Cr2O7(2-) + 6 IO4(-) + H+ (balance atoms whose oxidation number change)
31 H2O + 2 CrI3 => Cr2O7(2-) + 6 IO4(-) + H+ (balance O with H2O)
31 H2O + 2 CrI3 => Cr2O7(2-) + 6 IO4(-) + 62 H+ (balance H with H+)
31 H2O + 2 CrI3 => Cr2O7(2-) + 6 IO4(-) + 62 H+ + 54 e- (balance charge with electrons)
3. The reduction half-reaction is
Cl2 => Cl-
Cl2 => 2 Cl- (balance atoms whose oxidation number change)
2e- + Cl2 => 2 Cl- (balance charge with electrons)
27(2e- + Cl2 => 2 Cl-) (make electron gain equal electron loss)
54 e- + 27 Cl2 => 54 Cl-
4. Combine the two half-reactions
27 Cl2 + 31 H2O + 2 CrI3 => Cr2O7(2-) + 6 IO4(-) + 62 H+ + 54 Cl-
5. Add 62 OH- to each side (to neutralize H+ and make a basic solution)
62 OH- + 27 Cl2 + 31 H2O + 2 CrI3 => Cr2O7(2-) + 6 IO4(-) + 62 H+ + 62 OH- + 54 Cl-
62 OH- + 27 Cl2 + 31 H2O + 2 CrI3 => Cr2O7(2-) + 6 IO4(-) + 62 H2O + 54 Cl- (H+ + OH- => H2O)
62 OH- + 27 Cl2 + 2 CrI3 => Cr2O7(2-) + 6 IO4(-) + 31 H2O + 54 Cl- (simplify)
6. Add 62 K+ to each side (to recreate KOH, K2Cr2O7, KIO4, and KCl)
62 K+ + 62 OH- + 27 Cl2 + 2 CrI3 => 2 K+ + Cr2O7(2-) + 6 K+ + 6 IO4(-) + 31 H2O + 54 K+ + 54 Cl-
62 KOH + 27 Cl2 + 2 CrI3 => K2Cr2O7 + 6 KIO4 + 31 H2O + 54 KCl (combine ions)
Same as above. I think the half-reaction method is easier in this case, but you can decide…
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