Hello, I am having trouble balancing this redox reaction WITHOUT using the half-reaction method. I believe this is because not all of the elements are in (aq).
BrO3-(aq) + N2H4(g) --> Br-(aq) + N2(g)
Br is reduced by a factor of 6.
N is oxidized by a factor of 2.
I will include these factors as coefficients and attempt to balance the hydrogen and oxygen.
2BrO3-(aq) + 6N2H4(g) --> 2Br-(aq) + 6N2(g)+ 6H2O(l)
It seems like I should be adding 12H+ to the products side to balance the hydrogen, but this doesn't work. I know the answer is:
2BrO3-(aq) + 3N2H4(g) --> 2Br-(aq) + 3N2(g)+ 6H2O(l)
But why are my stoichiometric factors incorrect for N2H4 and N2?
Thank you!
The FIRST problem is that the loss of electrons for N2 is 4. N is +4 for the 2 N atoms and it goes to zero for the 2N atoms on the right. EACH N is -2 but you aren't dealing with EACH N. You are dealing with both on the left and both on the right.
Second thing, not enough to start World War III over, but you should not say that Br changes by a factor of 6. It changes by 6 electrons, yes, and its a GAIN of 6e but not a FACTOR of 6.
Br gains 6 electrons, both N change by 4 electrons. Multiply N by 6 and Br by 4 and see if things get easier.
The stoichiometric factors you provided for N2H4 and N2 are incorrect because they do not accurately reflect the changes in oxidation numbers of nitrogen (N) during the reaction.
In the given reaction, nitrogen in N2H4 is oxidized from a oxidation state of -2 in N2H4 to 0 in N2 gas. This change corresponds to an oxidation factor of 2.
Therefore, you should adjust the stoichiometric factor for N2H4 to account for this change:
BrO3-(aq) + 3N2H4(g) -> 2Br-(aq) + 3N2(g) + 6H2O(l)
Now, if we balance the hydrogen and oxygen by adding H2O and H+ ions, the balanced equation becomes:
BrO3-(aq) + 3N2H4(g) + 3H2O(l) -> 2Br-(aq) + 3N2(g) + 6H2O(l)
We can see that the balanced equation now satisfies both mass and charge balance, with the appropriate stoichiometric factors for N2H4 and N2 as determined by the change in oxidation state.
To balance a redox reaction without using the half-reaction method, you need to follow some steps. Let's go through them together and understand where the issue with your stoichiometric factors lies.
Step 1: Assign oxidation numbers to all elements in the reaction.
For BrO3-(aq):
Br: +5
O: -2
For N2H4(g):
N: -2
H: +1
For Br-(aq):
Br: -1
For N2(g):
N: 0
Step 2: Identify the elements that are undergoing oxidation and reduction.
In this reaction:
Br is reduced by a factor of 6 (from +5 to -1).
N is oxidized by a factor of 2 (from -2 to 0).
Step 3: Write the balanced skeletal equation by including the stoichiometric factors for the elements undergoing oxidation and reduction.
Your initial attempt to balance the equation with stoichiometric factors of 6 for Br and 2 for N is incorrect. This is because the stoichiometric factors should represent the ratio of moles of the substances involved in the reaction, not the change in oxidation number.
Let's adjust the stoichiometric factors based on the oxidation and reduction factors we identified:
BrO3-(aq) + 3N2H4(g) --> Br-(aq) + 3N2(g)
Now, let's try to balance the hydrogen and oxygen.
Oxygen: We have 3 oxygen atoms on the left (from BrO3-) and none on the right. Adding 3 H2O molecules to the right side will balance the oxygen.
BrO3-(aq) + 3N2H4(g) --> Br-(aq) + 3N2(g) + 3H2O(l)
Hydrogen: We have 12 hydrogen atoms on the left (from N2H4) and 6 on the right (from the water molecules). Adding 6 H+ ions to the left side will balance the hydrogen.
2BrO3-(aq) + 6N2H4(g) + 6H+ --> 2Br-(aq) + 6N2(g) + 3H2O(l)
Now the equation is balanced with respect to hydrogen and oxygen, and the stoichiometric factors for N2H4 and N2 are correctly adjusted based on the oxidation and reduction factors.
I hope this clarifies the issue and helps you understand how to balance redox reactions without using the half-reaction method.