how to balance chemical equations with redox reaction Cr(OH)3 + IO3{-} + OH{-} = CrO4{-2} + I{-}
http://chem.libretexts.org/Core/Analytical_Chemistry/Electrochemistry/Redox_Chemistry/Balancing_Redox_reactions
Go through these examples. Use the REDOX rules (all 7 of them).
7OH- + 5Cr(OH)3 + 3IO3- => 5CrO4-2 + 3I- + 11H2O
To balance the chemical equation for the redox reaction involving Cr(OH)3, IO3^-, OH^-, CrO4^2-, and I^-, follow the steps below:
Step 1: Assign oxidation numbers
Start by assigning oxidation numbers to each element in the reaction. The general rules to assign oxidation numbers are as follows:
- The oxidation number of any element by itself is zero.
- The sum of the oxidation numbers in a neutral compound is zero.
- The oxidation number of a monatomic ion is equal to its charge.
- The more electronegative element in a binary compound is assigned a negative oxidation number equal to the charge it would have as an ion.
- The oxidation number of hydrogen is +1 unless it is in a hydride compound where it is -1.
- The oxidation number of oxygen is usually -2 unless it is in a peroxide compound (such as H2O2) where it is -1.
Using these rules, the oxidation numbers for the given compounds are:
Cr(OH)3: Cr = +3
IO3^-: I = +5, O = -2
OH^-: O = -2, H = +1
CrO4^2-: Cr = +6, O = -2
I^-: I = -1
Step 2: Identify the species being oxidized and reduced
In this case, Cr is being reduced from +6 to +3, while I is being oxidized from -1 to +5.
The species being oxidized is called the reducing agent, and the species being reduced is called the oxidizing agent.
Step 3: Write the half-reactions
Separate the reaction into two half-reactions: the oxidation half-reaction and the reduction half-reaction.
Oxidation half-reaction: I^- -> IO3^-
Reduction half-reaction: Cr(OH)3 -> CrO4^2-
Step 4: Balance the atoms other than oxygen and hydrogen
Next, balance the atoms in each half-reaction, excluding hydrogen (H) and oxygen (O).
For the oxidation half-reaction:
I^- -> IO3^-
Add 6 I^- to the left side and 6 I^- to the right side to balance the iodine atoms.
6 I^- + H2O -> IO3^- + 6e^-
For the reduction half-reaction:
Cr(OH)3 -> CrO4^2-
Balance the chromium (Cr) atoms by adding a coefficient of 2 in front of Cr(OH)3.
2 Cr(OH)3 -> CrO4^2- + 8 H2O
Step 5: Balance oxygen by adding water molecules
To balance the oxygen (O), add water (H2O) molecules as needed to the side that lacks oxygen. Count the number of oxygen atoms on each side of the equation and adjust the coefficients of H2O accordingly.
For the oxidation half-reaction:
6 I^- + H2O -> IO3^- + 6e-
Add 6 H2O on the left side to balance the oxygen atoms.
6 I^- + 6 H2O -> IO3^- + 6e-
For the reduction half-reaction:
2 Cr(OH)3 -> CrO4^2- + 8 H2O
Add 4 H2O on the right side to balance the oxygen atoms.
2 Cr(OH)3 -> CrO4^2- + 8 H2O
Step 6: Balance hydrogen by adding hydrogen ions (H+)
To balance the hydrogen (H), add hydrogen ions (H+) as needed to the side that lacks hydrogen.
For the oxidation half-reaction:
6 I^- + 6 H2O -> IO3^- + 6e-
Add 12 H+ on the right side to balance the hydrogen atoms.
6 I^- + 6 H2O -> IO3^- + 6e- + 12 H+
For the reduction half-reaction:
2 Cr(OH)3 -> CrO4^2- + 8 H2O
Add 16 H+ on the left side to balance the hydrogen atoms.
2 Cr(OH)3 + 16 H+ -> CrO4^2- + 8 H2O
Step 7: Balance the charges using electrons
To balance the charges, add electrons (e-) to the side that has the more positive or less negative total charge.
For the oxidation half-reaction:
6 I^- + 6 H2O -> IO3^- + 6e-
The total charge on the left side is -6, while the total charge on the right side is -1. To balance the charges, add 5 electrons (e-) to the left side.
6 I^- + 6 H2O + 5e- -> IO3^-
For the reduction half-reaction:
2 Cr(OH)3 + 16 H+ -> CrO4^2- + 8 H2O
The total charge on the left side is +16, while the total charge on the right side is +2. To balance the charges, add 14 electrons (e-) to the right side.
2 Cr(OH)3 + 16 H+ + 14e- -> CrO4^2- + 8 H2O
Step 8: Balance the electrons
Equalize the number of electrons in both half-reactions by multiplying each half-reaction by a coefficient.
Multiply the oxidation half-reaction by 2 and the reduction half-reaction by 3 to balance the electrons.
12 I^- + 12 H2O + 10e- -> 2 IO3^-
6 Cr(OH)3 + 48 H+ + 42e- -> 3 CrO4^2- + 24 H2O
Step 9: Combine the half-reactions
Add the two half-reactions together and cancel out the common species on both sides of the equation.
Combine the two half-reactions:
12 I^- + 12 H2O + 10e- + 6 Cr(OH)3 + 48 H+ + 42e- -> 2 IO3^- + 3 CrO4^2- + 24 H2O
Cancel out the electrons (e-):
12 I^- + 12 H2O + 6 Cr(OH)3 + 48 H+ -> 2 IO3^- + 3 CrO4^2- + 24 H2O
Step 10: Finalize the balanced equation
Finally, simplify the equation and balance the number of water (H2O) molecules by adjusting coefficients if necessary.
12 I^- + 6 Cr(OH)3 + 48 H+ -> 2 IO3^- + 3 CrO4^2- + 36 H2O
And that's the balanced equation for the redox reaction Cr(OH)3 + IO3^- + OH^- = CrO4^2- + I^-!
To balance this chemical equation, we will follow these steps:
Step 1: Assign oxidation numbers to each element in the equation.
Step 2: Identify which elements are oxidized and which are reduced.
Step 3: Balance the atoms involved in the oxidation and reduction half-reactions separately.
Step 4: Balance the number of electrons transferred in the half-reactions.
Step 5: Combine the half-reactions and balance the final equation.
Let's start by assigning oxidation numbers:
The oxidation number of Cr (chromium) is +3 in Cr(OH)3 and +6 in CrO4{-2}.
The oxidation number of O (oxygen) is -2.
The oxidation number of H (hydrogen) is +1.
The oxidation number of I (iodine) is -1 in I{-} and +5 in IO3{-}.
The oxidation number of OH{-} (hydroxide) is -1.
Now, let's identify the oxidation and reduction half-reactions:
Oxidation Half-Reaction:
Cr(OH)3 → CrO4{-2} (Chromium changes from +3 to +6)
Reduction Half-Reaction:
IO3{-} + OH{-} → I{-} (Iodine changes from +5 to -1)
Next, let's balance the atoms involved in the oxidation and reduction half-reactions:
Oxidation Half-Reaction:
2Cr(OH)3 → CrO4{-2} (2 chromium atoms on both sides)
Reduction Half-Reaction:
6IO3{-} + 6OH{-} → 6I{-} + 3H2O (Balance oxygen atoms with water molecules, balance hydrogen atoms with H+ ions)
Now, let's balance the number of electrons transferred in the half-reactions:
Oxidation Half-Reaction:
2Cr(OH)3 → CrO4{-2} + 6e{-} (6 electrons are transferred)
Reduction Half-Reaction:
6IO3{-} + 6OH{-} → 6I{-} + 3H2O + 6e{-} (6 electrons are transferred)
Finally, combine the half-reactions, making sure the number of electrons on both sides are equal:
2Cr(OH)3 + 6IO3{-} + 14OH{-} → 2CrO4{-2} + 6I{-} + 8H2O
The balanced equation for the redox reaction is:
2Cr(OH)3 + 6IO3{-} + 14OH{-} → 2CrO4{-2} + 6I{-} + 8H2O