Hello, I am trying to balance this redox reaction WITHOUT using the half-reaction method. Here is the unbalanced equation:
MnO4 -(aq) + Zn(s)--> Mn 2+(aq) + Zn 2+(aq)
It looks as if both elements are being oxidized. The answer is supposed to be:
2MnO4 -(aq) + 16H +(aq) + 5Zn(s) --> 2Mn 2+(aq) + 8H2O(l) + 5Zn 2+(aq)
Thank you!
Mn goes from +7 on the left to +2 on the right so it is gaining 5e which is reduction. Zn is zero on the left and +2 on the right which is a loss of 2e which is oxidation. I helped you do one yesterday. I assume the corrections I made made it much easier to balance. Let me know if this one gives you any problems.
thank you so much! no, I've been doing many questions on half cells and cell potential and I mindlessly forgot to assign oxidation numbers, just going by the sign of the element as you would with half cell questions. you're awesome help! thanks!
To balance this redox reaction without using the half-reaction method, you can follow these steps:
1. Balance all atoms except hydrogen and oxygen. In this case, there is only one atom each of manganese (Mn) and zinc (Zn) on both sides of the equation, so they are already balanced.
MnO4^-(aq) + Zn(s) --> Mn^2+(aq) + Zn^2+(aq)
2. Balance oxygen atoms by adding water (H2O) molecules to the side that needs more oxygen. In this case, the left side needs oxygen.
MnO4^-(aq) + Zn(s) --> Mn^2+(aq) + Zn^2+(aq) + H2O(l)
3. Count the number of hydrogen atoms on both sides and balance them by adding hydrogen ions (H+) to the side that needs more hydrogen. In this case, the right side needs hydrogen.
MnO4^-(aq) + Zn(s) --> Mn^2+(aq) + Zn^2+(aq) + H2O(l) + H+(aq)
4. Balance the charges on both sides by adding electrons (e^-) to the side that needs more negative charge (reduction) and subtracting electrons from the side that needs less negative charge (oxidation). In this case, both Mn and Zn are being oxidized, so we will balance their charges.
MnO4^-(aq) + 5e^- + Zn(s) --> Mn^2+(aq) + Zn^2+(aq) + H2O(l) + H+(aq)
5. Check to make sure all atoms and charges are balanced. In this case, we have 1 Mn, 1 Zn, 4 O, 2 H, and no charge on both sides.
MnO4^-(aq) + 5e^- + Zn(s) --> Mn^2+(aq) + Zn^2+(aq) + 4H2O(l) + 4H+(aq)
6. Finally, multiply the entire equation by the appropriate factor if needed to ensure that all coefficients are whole numbers. In this case, multiplying by 2 will balance the Mn and Zn atoms.
2MnO4^-(aq) + 10e^- + 5Zn(s) --> 2Mn^2+(aq) + 5Zn^2+(aq) + 8H2O(l) + 8H+(aq)
Therefore, the balanced redox reaction without using the half-reaction method is:
2MnO4^-(aq) + 10e^- + 5Zn(s) --> 2Mn^2+(aq) + 5Zn^2+(aq) + 8H2O(l) + 8H+(aq)
To balance the redox reaction without using the half-reaction method, you can use the "balance by inspection" method. Here's how you can do it:
First, identify the elements that are changing oxidation states. In this case, it is manganese (Mn) and zinc (Zn).
Next, assign oxidation numbers to the different elements in the reaction. MnO4 - has an oxidation number of +7, while each O atom has an oxidation number of -2. Zn has an oxidation number of 0, and Mn 2+ and Zn 2+ both have oxidation numbers of +2.
Now, consider the change in oxidation numbers for each element. From MnO4 - to Mn 2+, the Mn atom decreases its oxidation number from +7 to +2 (a reduction process). From Zn to Zn 2+, the Zn atom increases its oxidation number from 0 to +2 (an oxidation process).
To balance the equation, start with the element that undergoes oxidation, which is zinc. There are five Zn atoms on the left side and only one on the right side, so you will need to multiply the Zn(s) reactant by 5 to have the same number of Zn atoms on both sides.
Next, move on to balancing the manganese atoms. In this case, you have two Mn atoms on the right side, but none on the left side. To balance this, you will need to add two MnO4 - ions on the left side.
Now, you need to balance the oxygen atoms. Each MnO4 - ion contains four O atoms, so by adding two MnO4 - ions on the left side, you have a total of eight O atoms. To balance this, you will need to add eight H2O molecules on the right side.
Finally, balance the hydrogen atoms. On the right side, you have 16 H+ ions from the H2O molecules. To balance this, you need to add 16 H+ ions on the left side.
The final balanced equation is:
2MnO4 -(aq) + 16H +(aq) + 5Zn(s) --> 2Mn 2+(aq) + 8H2O(l) + 5Zn 2+(aq)
Remember, this method is a visual approach that relies on trial and error. The half-reaction method is a more systematic and rigorous way to balance redox reactions.