1. In the balanced oxidation-reduction reaction (occurring in acidic condition) the coefficients before H+ and Zn2+
are ____ and ____, respectively?
Zn + NO3¯ → NH4+ + Zn2+ (unbalanced)
A. 4, 2
B. 4, 1
C. 6, 1
D. 10, 4
E. 12, 2
Anything you need to know about balancing redox equations is here.
http://www.chemteam.info/Redox/Redox.html
In this equation, Zn goes from zero to +2. N goes from +5 to -3.
To balance the oxidation-reduction reaction in acidic conditions, you need to follow these steps:
1. Start by assigning oxidation numbers to each element in the reaction.
Zn: 0
NO3¯: -1
NH4+: +1
Zn2+: +2
2. Identify the element undergoing oxidation and the element undergoing reduction.
In this case, Zn is being oxidized from an oxidation state of 0 to +2, and NO3¯ is being reduced from -1 to +1.
3. Write the half-reactions for both the oxidation and reduction processes.
Oxidation: Zn → Zn2+ + 2e^-
Reduction: NO3¯ + 3H+ + 2e^- → NH4+
4. Balance the atoms on each side of the half-reactions (excluding H and O atoms).
Oxidation: Zn → Zn2+ + 2e^-
Reduction: NO3¯ + 3H+ + 2e^- → NH4+
5. Balance the charge by adding electrons (e^-) to the appropriate side of each half-reaction.
Oxidation: Zn → Zn2+ + 2e^-
Reduction: NO3¯ + 3H+ + 2e^- → NH4+
6. Balance the number of electrons transferred in each half-reaction so that they are equal.
Oxidation: Zn → Zn2+ + 2e^-
Reduction: NO3¯ + 3H+ + 2e^- → NH4+
To balance the electrons, we need to multiply the oxidation half-reaction by 2 and the reduction half-reaction by 1.
7. Rewrite the balanced half-reactions.
Oxidation: 2Zn → 2Zn2+ + 4e^-
Reduction: 2NO3¯ + 6H+ + 4e^- → 2NH4+
8. To balance the number of hydrogen (H) atoms, add H+ ions to the appropriate side of the equation.
Oxidation: 2Zn → 2Zn2+ + 4e^-
Reduction: 2NO3¯ + 6H+ + 4e^- → 2NH4+
9. Finally, we determine the coefficients for H+ and Zn2+ in the balanced reaction.
The coefficients before H+ is 6 and before Zn2+ is 2.
Therefore, the correct answer is C. 6, 1
To determine the coefficients before H+ and Zn2+ in the balanced oxidation-reduction reaction, we need to follow a few steps:
Step 1: Determine the oxidation states.
In this equation, Zn starts with an oxidation state of 0, and NO3¯ has an oxidation state of -1.
Step 2: Identify the elements undergoing an oxidation or reduction.
In this equation, Zn is being oxidized from an oxidation state of 0 to +2, while NO3¯ is being reduced from an oxidation state of -1 to NH4+ with a total oxidation state of +1.
Step 3: Write half-reactions.
Separate the equation into two separate half-reactions, one for the oxidation and one for the reduction.
Oxidation half-reaction: Zn → Zn2+ + 2e^-
Reduction half-reaction: NO3¯ + 3H+ + 2e^- → NH4+
Step 4: Balance the half-reactions.
Balance the number of atoms and the charge on both sides of each half-reaction.
For the oxidation half-reaction, since there is only one Zn atom on both sides and the charge is already balanced, no further balancing is needed.
For the reduction half-reaction, there are three H+ on the right side, so we need to add 3 H+ ions on the left side to balance the hydrogen. Additionally, there are two electrons on the right side, so we need to add 2 electrons on the left side.
Balanced reduction half-reaction: NO3¯ + 3H+ + 2e^- → NH4+
Step 5: Combine the half-reactions.
To balance the number of electrons, multiply the oxidation half-reaction by 2 and the reduction half-reaction by 1.
2(Zn → Zn2+ + 2e^-)
NO3¯ + 3H+ + 2e^- → NH4+
Now, add the two half-reactions together, making sure that the electrons cancel out.
2Zn + NO3¯ + 3H+ → NH4+ + Zn2+
The coefficients before H+ and Zn2+ are 3 and 1, respectively.
Therefore, the correct answer is C. 6, 1.