The redox reaction below occurs in acidic solution. Balance it with the smallest whole number coefficients and determine the coefficient for H2O.
Zn(s) + NO3-(aq) ---> Zn2+(aq) + N2O(g)
i got 4 as the coefficient for water, but the answer says that it is 5
The Zn half is:
Zn ==> Zn^+2 + 2e
The NO3^- half is"
10H^+ + 8e + 2NO3^- ==> N2O + 5H2O
Multiply the Zn half by 4, the NO3^- half by 1 and add the two equations.
I have 5 for H2O also.
If you will type your equation in I will tell you why it is wrong. I know you can look at your answer and compare it with mine and see where the coefficients are not the same; however, it is possible to balance the equation atom for atom and even electron for electron and NOT be balanced.
Well, let's see if we can figure this out together!
First, let's balance the atoms that are not involved in oxidation or reduction. We have one zinc atom on each side, so that's already balanced.
Next, let's balance the nitrate ion (NO3-) by adding three more on the left side:
Zn(s) + 3NO3-(aq) ---> Zn2+(aq) + N2O(g)
Now, let's balance the oxygen atoms. We have 6 oxygen atoms on the right side, so we need to add 6 water molecules (H2O) on the left side:
Zn(s) + 3NO3-(aq) + 6H2O(l) ---> Zn2+(aq) + N2O(g)
Finally, let's balance the hydrogen atoms. We have 12 hydrogen atoms on the left side, so we need to add 12 protons (H+) on the right side. We can do this by adding 6 H+ ions on the right side:
Zn(s) + 3NO3-(aq) + 6H2O(l) ---> Zn2+(aq) + 2N2O(g) + 6H+(aq)
And there you have it! The balanced equation with the smallest whole number coefficients shows that the coefficient for water (H2O) is indeed 6, not 5. Keep up the balancing act!
To balance the redox reaction, first, we need to identify the oxidation states of each element involved.
Zn(s) has an oxidation state of 0 since it is an element in its standard state.
NO3-(aq) has an oxidation state of -1 since the overall charge of NO3- is 1-.
Zn2+(aq) has an oxidation state of +2 since it has a 2+ charge.
N2O(g) has an overall oxidation state of 0 since it is an elemental compound.
Next, we need to balance the elements that undergo oxidation or reduction separately.
The oxidation half-reaction is:
Zn(s) → Zn2+(aq)
The reduction half-reaction is:
NO3-(aq) → N2O(g)
Balance the atoms in each half-reaction by adding coefficients as necessary:
In the oxidation half-reaction:
Zn(s) → Zn2+(aq)
The Zn atoms are balanced on both sides.
In the reduction half-reaction:
2NO3-(aq) → N2O(g)
The N and O atoms are balanced on both sides.
Next, balance the electron transfer in the half-reactions by multiplying one or both of the half-reactions by appropriate coefficients:
In the oxidation half-reaction:
To balance the charge, we need to add 2 electrons to the reactant side:
Zn(s) → Zn2+(aq) + 2e-
In the reduction half-reaction:
The number of electrons is already balanced.
Now, we can combine the two half-reactions and cancel out common species:
2Zn(s) + 10NO3-(aq) → 2Zn2+(aq) + N2O(g) + 10e-
To balance the number of electrons, we need to multiply the oxidation half-reaction by 5 and the reduction half-reaction by 10:
10Zn(s) + 50NO3-(aq) → 10Zn2+(aq) + 5N2O(g) + 50e-
Finally, cancel out the electrons and recheck the coefficients:
10Zn(s) + 50NO3-(aq) → 10Zn2+(aq) + 5N2O(g)
The balanced equation shows that the coefficient for H2O is 0. There is no H2O involved in this redox reaction.
To balance a redox reaction, follow these steps:
Step 1: Identify the oxidation states of each element.
In this reaction, the oxidation state of Zn changes from 0 to +2, and the oxidation state of N changes from +5 to +2.
Step 2: Split the reaction into two half-reactions: oxidation and reduction.
Oxidation half-reaction: Zn(s) → Zn^2+(aq)
Reduction half-reaction: NO3^-(aq) → N2O(g)
Step 3: Balance the atoms in each half-reaction (excluding H and O).
The atoms are already balanced in both half-reactions.
Step 4: Balance the charges in each half-reaction by adding electrons (e^-).
Oxidation half-reaction: Zn(s) → Zn^2+(aq) + 2e^-
Reduction half-reaction: 2NO3^-(aq) + 10e^- → N2O(g)
Step 5: Balance the number of electrons in the reactions.
Multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2 to equalize the number of electrons.
5Zn(s) → 5Zn^2+(aq) + 10e^-
10NO3^-(aq) + 50e^- → 5N2O(g)
Step 6: Add the balanced half-reactions together.
5Zn(s) + 10NO3^-(aq) → 5Zn^2+(aq) + 5N2O(g)
Step 7: Balance the atoms of H and O by adding H2O to the appropriate side of the equation.
In this case, there are 10 oxygen atoms on the left side and only 5 oxygen atoms on the right side. To balance them, add 5 H2O to the left side:
5Zn(s) + 10NO3^-(aq) + 5H2O(l) → 5Zn^2+(aq) + 5N2O(g)
So, the coefficient for H2O in this balanced redox reaction is 5.