Balance the following chemical equation using the oxidation number method:
K2Cr2O7 + SnCl2 + HCl → CrCl3 + SnCl4 + H2O + KCl
K +1 ; +1
Cr +6 ; +3 gained 3 electrons
Sn +2 ; +4 lost 2 electrons
Cl -1; -1
so the ratio of Cr to Sn is 2:3 so start out with these coefficents
K2Cr2O7 and 3SnCl2
That should get you there pretty quickly.
To balance the given chemical equation using the oxidation number method, follow these steps:
Step 1: Assign oxidation numbers to each element in the equation.
In K2Cr2O7,
- Oxygen (O) always has an oxidation number of -2.
- Potassium (K) has an oxidation number of +1.
- The sum of the oxidation numbers in K2Cr2O7 should be 0 since it is a neutral compound.
So, 2K + 2(+1) + Cr + 7(-2) = 0
2K + 2 + Cr - 14 = 0
2K + Cr - 12 = 0
In SnCl2,
- Chlorine (Cl) has an oxidation number of -1.
- The sum of the oxidation numbers in SnCl2 should be 0 since it is a neutral compound.
So, Sn + 2(-1) = 0
Sn - 2 = 0
Sn = +2
In HCl,
- Chlorine (Cl) has an oxidation number of -1.
- Hydrogen (H) has an oxidation number of +1.
- The sum of the oxidation numbers in HCl should be 0 since it is a neutral compound.
So, H + (-1) = 0
H - 1 = 0
H = +1
Now, let's assign oxidation numbers to the remaining elements.
In CrCl3, chlorine (Cl) has an oxidation number of -1. So, the oxidation number of Cr has to be +3 to balance out the charges.
In SnCl4, chlorine (Cl) has an oxidation number of -1. So, the oxidation number of Sn has to be +4 to balance out the charges.
Step 2: Identify the elements being oxidized and reduced.
From the assigned oxidation numbers, we can see that:
- Cr is being reduced from +6 to +3.
- Sn is being oxidized from +2 to +4.
Step 3: Balance the elements being oxidized and reduced.
There are two Cr atoms in K2Cr2O7, so we need to multiply CrCl3 by 2 to balance the number of Cr atoms.
K2Cr2O7 + SnCl2 + HCl → 2CrCl3 + SnCl4 + H2O + KCl
There is one Sn atom in SnCl2, so we don't need to multiply SnCl4.
K2Cr2O7 + SnCl2 + HCl → 2CrCl3 + SnCl4 + H2O + KCl
Step 4: Balance the remaining atoms.
Balance the number of Cl atoms by adding 14HCl to the right side of the equation.
K2Cr2O7 + SnCl2 + 14HCl → 2CrCl3 + SnCl4 + H2O + KCl
Balance the number of K atoms by adding 2KCl to the right side of the equation.
K2Cr2O7 + SnCl2 + 14HCl → 2CrCl3 + SnCl4 + H2O + 2KCl
Balance the number of O atoms by adding 7H2O to the right side of the equation.
K2Cr2O7 + SnCl2 + 14HCl → 2CrCl3 + SnCl4 + 7H2O + 2KCl
Therefore, the balanced chemical equation using the oxidation number method is:
K2Cr2O7 + SnCl2 + 14HCl → 2CrCl3 + SnCl4 + 7H2O + 2KCl
To balance this chemical equation using the oxidation number method, follow these steps:
Step 1: Assign oxidation numbers to all the elements in the equation.
In this equation, we need to determine the oxidation numbers for each element. Here are some common rules:
- The oxidation number of an element in its free state is always zero.
- The oxidation number of an alkali metal (Group 1) is always +1.
- The oxidation number of fluorine is always -1.
- The sum of the oxidation numbers in a compound is equal to the overall charge of the compound.
Using these rules, here are the assigned oxidation numbers:
K2Cr2O7 ⇒ K(+1), Cr(unknown), O(-2)
SnCl2 ⇒ Sn(unknown), Cl(-1)
HCl ⇒ H(+1), Cl(-1)
CrCl3 ⇒ Cr(unknown), Cl(-1)
SnCl4 ⇒ Sn(unknown), Cl(-1)
H2O ⇒ H(+1), O(-2)
KCl ⇒ K(+1), Cl(-1)
Step 2: Identify which elements are undergoing oxidation and reduction.
In this equation, Cr is being reduced from an oxidation state of +6 in K2Cr2O7 to an oxidation state of +3 in CrCl3, and Sn is being oxidized from an oxidation state of +2 in SnCl2 to an oxidation state of +4 in SnCl4.
Step 3: Determine the number of electrons transferred.
To balance the oxidation and reduction half-reactions, the number of electrons transferred in each half-reaction should be the same. In this case, Cr is gaining 3 electrons, while Sn is losing 2 electrons. Therefore, we need to multiply the Cr reduction half-reaction by 2 and the Sn oxidation half-reaction by 3 to equalize the number of electrons transferred.
Step 4: Balance the elements and charges.
Now, we can balance the elements and charges in the equation. Start by balancing all the non-hydrogen and non-oxygen elements first, then balance the oxygen atoms using water (H2O), and finally balance the hydrogen atoms using H+ ions.
The balanced equation using the oxidation number method is:
2K2Cr2O7 + 3SnCl2 + 14HCl → 2CrCl3 + 3SnCl4 + 7H2O + 4KCl