Balance the redox reaction of eq.
Cu2S + SO4(-2) gives CuO + SO2 (Acidic )
3Cu2s+8So4{-}+8H{+}=6Cuo+11So2+4H2o
2Cu2S + 3SO4^2- + 2H2O = 4CuO + 5SO2 + 4H+
Why did the copper and sulfur go to couples therapy? Because they needed help in balancing their redox reaction!
Let's break it down and find a solution:
First, let's look at the oxidation numbers of the elements involved:
Cu2S: Copper (Cu) has an oxidation number of +2, and sulfur (S) has an oxidation number of -2.
SO4(-2): Sulfur has an oxidation number of +6, and oxygen (O) has an oxidation number of -2. Since there are four oxygen atoms in the sulfate ion (SO4), the total charge will be -2.
CuO: Copper has an oxidation number of +2, and oxygen has an oxidation number of -2.
SO2: Sulfur has an oxidation number of +4, and oxygen has an oxidation number of -2.
Now, let's balance the equation step-by-step:
1. Balance the atoms other than oxygen and hydrogen:
Cu2S + SO4(-2) → CuO + SO2
2. Balance the sulfur atoms by adding a coefficient of 2 in front of SO2:
Cu2S + SO4(-2) → CuO + 2SO2
3. Balance the copper atoms by adding a coefficient of 2 in front of CuO:
Cu2S + SO4(-2) → 2CuO + 2SO2
4. Balance the oxygen atoms by adding water (H2O) molecules:
Cu2S + SO4(-2) → 2CuO + 2SO2 + 2H2O
5. Balance the hydrogen atoms by adding hydrogen ions (H+):
Cu2S + SO4(-2) + 4H+ → 2CuO + 2SO2 + 2H2O
And there you have it! The balanced equation is:
Cu2S + SO4(-2) + 4H+ → 2CuO + 2SO2 + 2H2O
The copper and sulfur are now in a better relationship, thanks to some balancing therapy!
To balance a redox reaction, follow these steps:
1. Write the skeleton equation:
Cu2S + SO4(-2) → CuO + SO2
2. Assign oxidation numbers to each element. In this reaction, the oxidation states of Cu, S, O, and S remain unchanged. However, the oxidation state of S in SO4(-2) is +6.
3. Separate the reaction into two half-reactions: the oxidation half-reaction and the reduction half-reaction.
Oxidation half-reaction: Cu2S → CuO
Reduction half-reaction: SO4(-2) → SO2
4. Balance each half-reaction:
Oxidation half-reaction:
The element S has a different oxidation state on each side of the equation. Since the change is from -2 to +4, it needs to gain 6 electrons to balance the charge. Multiply the Cu2S by 6 to achieve this:
6Cu2S → 6CuO
Reduction half-reaction:
The element S experiences no change in its oxidation state. Balancing the number of S atoms on both sides, we have:
SO4(-2) → SO2
5. Balance the charge in each half-reaction by adding electrons:
Oxidation half-reaction:
6Cu2S → 6CuO + 6e-
Reduction half-reaction:
SO4(-2) + 6e- → SO2
6. Make the number of electrons equal in both half-reactions by multiplying the reduction half-reaction by 6:
6SO4(-2) + 36e- → 6SO2
7. Add the two balanced half-reactions together:
6Cu2S + 6SO4(-2) + 36e- → 6CuO + 6SO2 + 6e-
8. Combine the like terms and eliminate the electrons:
6Cu2S + 6SO4(-2) → 6CuO + 6SO2
9. Finally, balance the elements other than S and O (which are already balanced):
6Cu2S + 6SO4(-2) → 6CuO + 3SO2
The balanced redox equation for the reaction Cu2S + SO4(-2) → CuO + SO2 is 6Cu2S + 6SO4(-2) → 6CuO + 3SO2 in an acidic solution.
2Cu2S + SO4^2- + 2H2O = 4CuO + 5SO2 + 4H+
3 Cu2S + 8 SO4{-}2 + 8 H{+} = 6 CuO + 11 SO2 + 4 H2O
First step.
Cu2S ==> 2CuO
Cu changes from total of +2 on the left to total of +4 on the right.
S changes from +6 in SO4^2- to +4 in SO2.
Now you balance the two half reactions.