Chemistry: Balancing Redox reactions

In summary, the conversation is about balancing a redox reaction occurring in acidic solution. The correct balanced equation, including phases, is 5Sn(s) + 2MnO4-(aq) + 16H+(aq) → 5Sn2+(aq) + 2Mn2+(aq) + 8H2O. However, there was a question about the placement of subscripts and super subscripts, which may have been resolved by correcting the placement of MnO-4 to MnO4-.
  • #1
needOfHelpCMath
72
0
What do i have wrong here for my chemical equation? Seems to be right but my homework won't accept it.

Balance the redox reaction occurring in acidic solution.

Sn(s)+MnO−4(aq) → Sn2+(aq)+Mn2+(aq)

Express your answer as a chemical equation including phases.
5Sn+2MnO−4+16H+→5Sn2++2Mn2++8H2O


is this correct?
 
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  • #2
needOfHelpCMath said:
What do i have wrong here for my chemical equation? Seems to be right but my homework won't accept it.

Balance the redox reaction occurring in acidic solution.

Sn(s)+MnO−4(aq) → Sn2+(aq)+Mn2+(aq)

Express your answer as a chemical equation including phases.
5Sn+2MnO−4+16H+→5Sn2++2Mn2++8H2O


is this correct?

Hi needOfHelpCMath!

It seems correct to me, although the question does ask to include phases.
So should it perhaps be:
$$Sn(s)+MnO^−_4(aq) → Sn^{2+}(aq)+Mn^{2+}(aq)$$
$$5Sn(s)+2MnO^−_4(aq) + 16H^+(aq)→ 5Sn^{2+}(aq)+2Mn^{2+}(aq) + 8H_2O
$$
(Wondering)
 
  • #3
I like Serena said:
Hi needOfHelpCMath!

It seems correct to me, although the question does ask to include phases.
So should it perhaps be:
$$Sn(s)+MnO^−_4(aq) → Sn^{2+}(aq)+Mn^{2+}(aq)$$
$$5Sn(s)+2MnO^−_4(aq) + 16H^+(aq)→ 5Sn^{2+}(aq)+2Mn^{2+}(aq) + 8H_2O
$$
(Wondering)

So inputed your answer and it said "Check your placement of subscripts and super subscripts"
 
  • #4
needOfHelpCMath said:
So inputed your answer and it said "Check your placement of subscripts and super subscripts"

Perhaps MnO-4 should be the other way around. That is, it should be MnO4-, since it's really $(MnO_4)^-$. (Thinking)
 

What is a redox reaction?

A redox reaction is a type of chemical reaction in which there is a transfer of electrons between two or more substances. One substance loses electrons (oxidation) while another substance gains electrons (reduction).

Why is it important to balance redox reactions?

Balancing redox reactions is important because it ensures that the same number of electrons are transferred from one substance to another. This is necessary in order to accurately represent the reaction and calculate the correct stoichiometric coefficients.

What are the steps for balancing a redox reaction?

The steps for balancing a redox reaction are as follows:

  1. Identify the oxidation numbers for each element in the reaction.
  2. Determine which elements are being oxidized and reduced.
  3. Write separate half-reactions for the oxidation and reduction processes.
  4. Balance the number of atoms for each element in each half-reaction.
  5. Balance the charge on each half-reaction by adding electrons to one side.
  6. Multiply the half-reactions by the appropriate coefficients to equalize the number of electrons transferred.
  7. Add the half-reactions together, canceling out any common terms.
  8. Double-check that the number of atoms and the charge are balanced on both sides of the equation.

How do you know if a redox reaction is balanced?

A redox reaction is considered balanced when the number of atoms and the charge are equal on both sides of the equation. This means that the same amount of electrons are transferred from one substance to another.

Are there any shortcuts for balancing redox reactions?

Yes, there are several shortcuts for balancing redox reactions. One method is the half-reaction method, where the oxidation and reduction half-reactions are balanced separately and then combined. Another method is the oxidation number method, which involves assigning oxidation numbers to each element and using these numbers to balance the reaction. Additionally, there are online calculators and software programs available that can help balance redox reactions.

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