Balancing Equations with Half-Reaction and Oxidation Number Methods

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Discussion Overview

The discussion revolves around balancing chemical equations using the half-reaction and oxidation number methods. Participants are specifically addressing two equations involving cobalt and manganese, as well as copper and nitric acid, exploring the steps and reasoning involved in the balancing process.

Discussion Character

  • Homework-related
  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • A participant expresses uncertainty about balancing the first equation involving cobalt and manganese, while feeling more confident about the second equation with copper.
  • Another participant points out that the original poster's answer for the second equation is incomplete, emphasizing the need to show the work involved in balancing.
  • A participant provides a breakdown of the oxidation states and electron transfers for the second equation, indicating the changes in oxidation states for copper and nitrogen.
  • There is a suggestion to use TexAide for typing chemical symbols and a discussion about the oxidation states of manganese and cobalt.
  • One participant corrects the terminology regarding manganese, suggesting the term "manganous" instead of "manganic."
  • Another participant outlines the need to balance the changes in electrons for both half-reactions and suggests multiplying the half-reactions to achieve balance.
  • A later reply confirms understanding of the concepts discussed, indicating progress in the participant's comprehension.

Areas of Agreement / Disagreement

Participants express varying levels of understanding and confidence regarding the balancing methods. There is no consensus on the final balanced equations, and multiple approaches and corrections are presented throughout the discussion.

Contextual Notes

Some participants mention difficulties in typing chemical equations, which may affect the clarity of their contributions. The discussion includes incomplete steps in the balancing process, and there are unresolved aspects regarding the balancing of oxygen and hydrogen in the equations.

Who May Find This Useful

This discussion may be useful for students learning about balancing chemical equations, particularly those interested in the half-reaction and oxidation number methods.

yoshi6
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Hello, I am new at this, I am not exactly sure how this works. I have a question where I have to balance two equations using either the half-reaction or the oxidation number method... they are as follows.:

1. Co + MnO4- + H+ = Co+2 + Mn+2 + H2O

2. Cu + HNO3 = Cu(NO3)2 + NO + H2O

I am having trouble with the first one. I think I get the second one okay; I am not sure how to type the whole process so here is my final answer for 2;

3Cu + 8HNO3 -> 3Cu(NO3)2 + 2NO + 4H2O

Thankyou
 
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thanks...umm...is my second one right? If you don't mind checking?
 
Your answer is incomplete. You were asked to balance the equations using one of the two methods. That implies that you are to show your work not just the final answer. The equation is mass balanced as you have shown it.
 
okay I hope this makes sense:

0 +5 +2 +2
Cu + HNO3 ---> Cu(NO3)2 + NO + H20

0 -2e +2
Cu --------> Cu(-2e) x 3 = -6e

+5 +3e +2
N-----------> N(+3e) x 2 = +6e

3Cu + 2HNO3 ---> 3Cu(NO3)2 + 2NO + H2O (incomplete)
3Cu + 8HNO3-----> 3Cu(NO3)2 + 2NO + H2O (incomplete)

3Cu + 8HNO3-----> 3Cu(NO3)2 + 2NO + 4H2O
 
it didn't work out as well as I hoped. I find it difficult to type these kinds of equations. Sorry.
 
yoshi6 wants help with:
1. Co + MnO4- + H+ = Co+2 + Mn+2 + H2O

For the "typing" of the symbolisms, try using TexAide; but for the analysis to balance the reaction equation, use this information:
Cobalt changed its state from 0 to +2.
The manganate changed to become a +2 Manganese (not sure if can call "manganic" ion);
Each oxygen carries -2, and there are 4 oxygens on permanganate, but the permanganate net charge is -1 (as you have indicated). So you want to find the state for the permanganate:
-1 = x + 4(-2)
x = +7

So, next, balance the change in electrons for each half reaction between Co and MnO4-;
Co looses 2 electrons; Mn goes from +7 to +2, so it gained 5 electrons. You must balance the changes in electrons.
For the Co half, multiply by 5; for the Mn half, multiply by 2.

Maybe someone else could explain the balancing of oxygens and water and either OH- or H+, or maybe you can balance that part yourself.
 
Co + Mn^{7+}O_4^- + H^+ \rightarrow Co^{2+} + Mn^{2+} + H_2O

First we have to find which species is being reduced, and which is being oxidized. through this we'll know which species is gaining/losing electrons.

Co^{0} + Mn^{+7} \rightarrow Co^{+2} + Mn^{+2}

Co loses 2 electrons, Mn gains 5 electrons. balance Co and Mn so that they have equal amount of electrons, but that won't be your final step.
 
Last edited:
  • #10
Thank you I get it now
 
  • #11
where is your new question?
 
  • #12
oh it is under math and science tutorials...I thought that was where I was supposed to put it? Maybe not...sorry
 

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