Balance the following redox reations:
K2Cr2O7 + Nal + H2SO4 --- Cr2(SO4)3 + I2 + H2O + Na2SO4 + K2SO4
To balance the given redox reaction, we need to ensure that the number of atoms and charges are equal on both sides of the reaction equation. Here's how to balance this equation step by step:
Step 1: Identify the oxidation states of each element:
In order to balance a redox reaction, it is essential to determine the oxidation states of the elements involved. The oxidation state of an element is a measure of the number of electrons it has gained or lost.
K2Cr2O7:
In this compound, the oxidation state of potassium (K) is always +1.
The sum of the oxidation states in K2Cr2O7 is: 2*(+1) + 2*x + 7*(-2) = 0
Simplifying the equation: 2 + 2x -14 = 0
2x -12 = 0
2x = 12
x = +6
Here, chromium (Cr) has an oxidation state of +6.
H2SO4:
Hydrogen (H) is generally +1 in compounds, and oxygen (O) is typically -2.
The sulfur (S) has an oxidation state of x.
Hence, 2*(+1) + x + 4*(-2) = 0
2 + x - 8 = 0
x - 6 = 0
x = +6
Here, sulfur (S) has an oxidation state of +6.
Nal:
Since sodium (Na) is always +1 and iodine (I) is generally -1, we can conclude that aluminum (Al) has an oxidation state of -1.
Cr2(SO4)3:
The sulfate anion (SO4) has a charge of -2. Therefore, each sulfur (S) atom has an oxidation state of +6 since there are three of them. The overall charge for this compound must be 0, so the oxidation state of chromium (Cr) must be +3.
I2:
Since iodine (I) is in its elemental form, its oxidation state is 0.
H2O:
For water (H2O), hydrogen (H) is always +1, and oxygen (O) is always -2. Thus, the H2O molecule is already balanced.
Na2SO4 and K2SO4:
Both compounds consist of ions, and the charges are already balanced. Thus, they don't need further adjustment.
Step 2: Determine the number of each element on both sides of the equation:
Count the number of atoms of each element on the reactant (left) and product (right) sides of the equation:
K2Cr2O7:
Reactant side: 2 K, 1 Cr, 7 O
Product side: 2 K, 2 Cr, 14 O
H2SO4:
Reactant side: 2 H, 1 S, 4 O
Product side: 2 H, 1 S, 4 O
Nal:
Reactant side: 2 Na, 1 I
Product side: 2 Na, 1 I
Cr2(SO4)3:
Reactant side: 2 Cr, 3 S, 12 O
Product side: 2 Cr, 3 S, 12 O
I2:
Reactant side: 1 I
Product side: 1 I
H2O:
Reactant side: 2 H, 1 O
Product side: 2 H, 1 O
Na2SO4:
Reactant side: 2 Na, 1 S, 4 O
Product side: 2 Na, 1 S, 4 O
K2SO4:
Reactant side: 2 K, 1 S, 4 O
Product side: 2 K, 1 S, 4 O
Step 3: Balance the equation:
Now, let's balance the equation by adjusting the coefficients of the compounds:
2 K2Cr2O7 + 10 Nal + 8 H2SO4 → 6 Cr2(SO4)3 + 21 I2 + 16 H2O + 10 Na2SO4 + 2 K2SO4
After balancing, both sides of the equation now have equal numbers of atoms and charges.
I don't want to simply balance it for you because you don't learn anything that way. Here are some hints.
Cr moves from +6 for each Cr atom on the left to +3 for each Cr atom on the right. I moves from -1 on the left to 0 on the right.
hint #2. First thing you do,
a. balance the Cr atoms on left and right and count total electrons transferred. As it is written, Cr is ok since there are two on the left and two on the right.
b. I: There is only one on the left and 2 on the right; therefore, place a 2NaI so I goes from -2 on the left to zero on the right.
Go from there.