I need to write the total ionic equation for this equation.
Copper(ii)sulfate + iron = iron(iii)+ copper.
Won't it be Cu2+ + SO4-2 + Fe = Fe3+ + Cu ?
The answer key say
3Cu2+ + 3SO4-2 + 2Fe = 2Fe3+ + 3SO4-2 + 3Cu
Where did they get the sulfate for the product from? Didn't the words equation say "iron(iii)+ copper."? Thanks
Your first statement(the words equation) is not correct because there is no sulfate on the product side. It should be
Copper(ii)sulfate + iron = iron(iii) sulfate + copper.
The world equation didn't have any sulfate on the product side :(. I'm guessing such formula can't be right since how can there be sulfate on one side and there none on the other. So the words equation is wrong right? Thanks
Yes the word equation is wrong. Just an omission in the text.
To write a balanced total ionic equation, we first need to understand the chemical formula of each compound involved and the charges of the ions present.
In this case, the chemical formula for copper(II) sulfate is CuSO4, which indicates that the Cu2+ ion is combined with the SO42- ion. The given equation: "Copper(II) sulfate + iron = iron(III) + copper" is not balanced and does not provide sufficient information to write the total ionic equation.
To properly balance the equation, we need to ensure that the number of atoms of each element and the net charge is the same on both sides. Here's how to do it:
1. Start by balancing the metal atoms.
Cu2+ + Fe = Fe3+ + Cu
2. Next, balance the non-metal atoms.
Cu2+ + Fe + 4O = Fe3+ + Cu + 4O
3. Finally, balance the charges by adjusting the number of ions.
3Cu2+ + 3(Fe2+) + 12O2- = 2(Fe3+) + 3Cu2+ + 12O2-
Since the Cu2+ and SO42- ions appear on both the reactant and product sides and do not undergo any changes, they can be considered as spectator ions and can be canceled out.
Thus, the balanced total ionic equation is:
3Fe2+ + 4O2- = 2Fe3+ + 2Fe2+ + 4O2-
Note that the charges and the number of atoms on both sides of the equation are balanced.