Balance the following redox (oxidation-reduction) reaction under basic conditions. (Include states-of-matter under the given conditions in your answer. Use the lowest possible whole number coefficients.)
CrO42−(aq) + S2O32−(aq) → Cr3+(aq) + SO42−(aq)
To balance the redox reaction under basic conditions, follow these steps:
Step 1: Identify the atoms being oxidized and reduced.
In this reaction, Chromium (Cr) is being reduced from a +6 oxidation state in CrO4^2- to a +3 oxidation state in Cr^3+, and Sulfur (S) is being oxidized from a +2 oxidation state in S2O3^2- to a +6 oxidation state in SO4^2-.
Step 2: Write the unbalanced redox half-reactions.
a) Reduction half-reaction:
CrO4^2- → Cr^3+
b) Oxidation half-reaction:
S2O3^2- → SO4^2-
Step 3: Balance the atoms in each half-reaction.
a) Reduction half-reaction:
CrO4^2- → Cr^3+
Cr: 1 Cr on both sides
O: 4 O on the left, 0 O on the right (add 4 OH- on the right to balance)
The reduction half-reaction becomes:
CrO4^2- + 4OH- → Cr^3+ + 4H2O
b) Oxidation half-reaction:
S2O3^2- → SO4^2-
S: 2 S on the left, 1 S on the right
O: 6 O on the left, 4 O on the right (add 2 H2O on the left to balance)
The oxidation half-reaction becomes:
S2O3^2- + 2H2O → SO4^2- + 4OH-
Step 4: Balance the charges in each half-reaction.
a) Reduction half-reaction:
CrO4^2- + 4OH- → Cr^3+ + 4H2O
Cr: +6 charge on the left, +3 charge on the right (add 3 electrons to the left)
The reduction half-reaction becomes:
CrO4^2- + 4OH- + 3e- → Cr^3+ + 4H2O
b) Oxidation half-reaction:
S2O3^2- + 2H2O → SO4^2- + 4OH-
S: +2 charge on the left, +6 charge on the right (add 4 electrons to the right)
The oxidation half-reaction becomes:
S2O3^2- + 2H2O → SO4^2- + 4OH- + 4e-
Step 5: Balance the electrons transferred.
To balance the number of electrons transferred, multiply the reduction half-reaction by 4 and the oxidation half-reaction by 3:
4(CrO4^2- + 4OH- + 3e-) → 4(Cr^3+ + 4H2O)
3(S2O3^2- + 2H2O → SO4^2- + 4OH- + 4e-)
Step 6: Combine the half-reactions and simplify.
12CrO4^2- + 16OH- + 12e- → 12Cr^3+ + 16H2O
12S2O3^2- + 24H2O → 4SO4^2- + 16OH- + 16e-
Step 7: Cancel out the common species.
Cancel out the 16OH- and 16e- on both sides:
12CrO4^2- + 12S2O3^2- + 24H2O → 12Cr^3+ + 4SO4^2- + 8H2O
Simplify the equation:
6CrO4^2- + 6S2O3^2- + 12H2O → 6Cr^3+ + 2SO4^2- + 4H2O
Step 8: Simplify the equation further.
6CrO4^2- + 6S2O3^2- + 8H2O → 6Cr^3+ + 2SO4^2-
To balance a redox reaction under basic conditions, you need to follow a series of steps. Here's a step-by-step explanation of how to balance the given reaction:
Step 1: Assign oxidation numbers
Start by assigning oxidation numbers to each element in the reaction. The oxidation number is a positive or negative value that represents the charge an atom would have if electrons were transferred completely.
CrO42-(aq): The oxygen atoms in CrO42- are typically assigned an oxidation number of -2. Since there are four oxygen atoms, the total oxidation number contributed by oxygen is -8.
CrO42- → Cr3+(aq): Therefore, the chromium (Cr) atom must have an oxidation number of +6 in order to balance out the total charge to -2.
S2O32-(aq): The two oxygen atoms in S2O32- each have an oxidation number of -2. Since there are three oxygen atoms, the total oxidation number contributed by oxygen is -6.
S2O32- → SO42-(aq): Therefore, the sulfur (S) atom must have an oxidation number of +6 in order to balance out the total charge to -2.
Step 2: Identify the elements undergoing oxidation and reduction
In this reaction, chromium (Cr) is being reduced because it changes from an oxidation state of +6 to +3, indicating a reduction (gaining electrons). Sulfur (S), on the other hand, is being oxidized because it changes from an oxidation state of +2 to +6, indicating an oxidation (losing electrons).
Step 3: Split the reaction into two half-reactions
Now, separate the reaction into two half-reactions - one for the oxidation and one for the reduction. The oxidation half-reaction involves the species being oxidized, while the reduction half-reaction involves the species being reduced.
Oxidation half-reaction: S2O32-(aq) → SO42-(aq) (oxidation)
Reduction half-reaction: CrO42-(aq) → Cr3+(aq) (reduction)
Step 4: Balance the atoms other than hydrogen and oxygen
Balance the atoms of each half-reaction, excluding hydrogen (H) and oxygen (O).
Oxidation half-reaction: Since there is one sulfur (S) atom on both sides, the sulfur is already balanced.
Reduction half-reaction: Since there is one chromium (Cr) atom on both sides, the chromium is already balanced.
Step 5: Balance the oxygen atoms
Balance the oxygen (O) atoms in each half-reaction by adding water (H2O) molecules to the side that requires oxygen atoms.
Oxidation half-reaction: S2O32-(aq) → SO42-(aq) + 2H2O(l) (oxidation)
Reduction half-reaction: CrO42-(aq) + 4H2O(l) → Cr3+(aq) (reduction)
Step 6: Balance the hydrogen atoms
Balance the hydrogen (H) atoms by adding hydrogen ions (H+) to the side that requires hydrogen atoms.
Oxidation half-reaction: 2H+ + S2O32-(aq) → SO42-(aq) + 2H2O(l) (oxidation)
Reduction half-reaction: CrO42-(aq) + 4H2O(l) → Cr3+(aq) + 4H+ (reduction)
Step 7: Balance the charges
Balance the charges on both sides of the half-reactions by adding electrons (e-) to the side that requires balance.
Oxidation half-reaction: 2e- + 2H+ + S2O32-(aq) → SO42-(aq) + 2H2O(l) (oxidation)
Reduction half-reaction: CrO42-(aq) + 4H2O(l) + 6e- → Cr3+(aq) + 4H+ (reduction)
Step 8: Equalize the number of electrons in both half-reactions
Multiply each half-reaction by the appropriate factor to equalize the number of electrons in both reactions. In this case, the oxidation half-reaction requires 6 electrons, while the reduction half-reaction provides 6 electrons. Multiplying the oxidation half-reaction by 6 will match the electrons in both reactions.
Oxidation half-reaction: 6e- + 6H+ + 3S2O32-(aq) → 3SO42-(aq) + 6H2O(l) (oxidation)
Reduction half-reaction: CrO42-(aq) + 4H2O(l) + 6e- → Cr3+(aq) + 4H+ (reduction)
Step 9: Combine the half-reactions
Add the two half-reactions together to obtain the balanced overall reaction. Make sure to cancel out species that are on both sides of the reaction equation.
6e- + 6H+ + 3S2O32-(aq) + CrO42-(aq) + 4H2O(l) → 3SO42-(aq) + Cr3+(aq) + 4H+ +6H2O(l)
Step 10: Simplify the overall reaction
Simplify the overall balanced reaction by canceling out species that appear on both sides of the reaction equation.
6H+ + 3S2O32-(aq) + CrO42-(aq) + 4H2O(l) → 3SO42-(aq) + Cr3+(aq) + 10H2O(l)
The balanced equation for the given redox reaction under basic conditions is:
6H+ + 3S2O32-(aq) + CrO42-(aq) + 4H2O(l) → 3SO42-(aq) + Cr3+(aq) + 10H2O(l)
Sure! Let's balance this redox reaction under basic conditions:
CrO42−(aq) + S2O32−(aq) → Cr3+(aq) + SO42−(aq)
First, let's balance the oxidation state of each element:
Cr: +6 → +3
S: +2 → +6
Next, let's balance the number of each element on the reactant and product side:
CrO42- + 3S2O32- → Cr3+ + 3SO42-
Now, let's balance the oxygens by adding water molecules to the appropriate side of the equation:
CrO42- + 3S2O32- + 4H2O→ Cr3+ + 3SO42-
Finally, let's balance the hydrogens by adding hydroxide ions (OH-) to the side that needs them:
CrO42- + 3S2O32- + 4H2O + 10OH-→ Cr3+ + 3SO42- + 6H2O
And there you have it! The balanced redox reaction under basic conditions, with each element and charge balanced nicely.
Here us a site that tells you the details.
http://www.chemteam.info/Redox/Redox.html
I don't do them that way. Here is how I do the CrO4^-. You can add the states.
CrO4^2- ==> Cr^3+
Cr is +6 on left and +3 on right. Add 3e to left.
CrO4^2- +3e ==> Cr^3+
Count the charge on the left (-5) and right (+3); add OH^- to the right to balance the charge.
CrO4^2- + 3e ==> Cr^3+ + 8OH^-
Now add H2O to balance.
CrO4^2- + 4H2O + 3e ==> Cr^3+ + 8OH^-