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# When balancing a redox reaction, i dont understand the part about adding electrons to balance the charge?

Topic: How to write a mass balance equation chemistry
June 16, 2019 / By Bess
Question: howdo you know how many electrons to add to the half reactions? it might be because i dont understand how to add up the charges in a chemical reaction. can someone please exaplin this? thank you so much.

## Best Answers: When balancing a redox reaction, i dont understand the part about adding electrons to balance the charge?

Afrika | 4 days ago
It's easier to explain with examples so I'll include one, if you don't mind. How would we balance the following redox reaction in acidic solution? HNO3 + H3AsO3 --> NO + H3AsO4 + H2O The first task is to determine what is being oxidized and what is being reduced. You have to assign oxidation numbers to all the elements in the equation and see what's changing from the left side to the right side. If you don't know how to assign oxidation numbers, go here: http://chemistry.about.com/od/generalche... When we do this we find that the oxidation number (hereafter, O.N.) of the atoms are as follows: In HNO3: H = +1 N = +5 O = -2 In H3AsO3: H = +1 As = +3 O = -2 In NO: N = +2 O = -2 In H3AsO4: H = +1 As = +5 O = -2 In H2O: H = +1 O = -2 With all the oxidation numbers assigned we see that arsenic is being oxidized from +3 to +5 while nitrogen is being reduced from +5 to +2. Let's start by writing the oxidation half-reaction. The half-reaction starts by showing all ions or molecules that contain the atom being oxidized. Acids can be treated as fully ionized. AsO3^-3 --> AsO4^-3 Now we need to add electrons to account for the change in O.N. of the arsenic atom. Oxidation involves LOSS of electrons, so we should put the electrons on the right side of the half-reaction. How many electrons? Since arsenic's O.N. increases from +3 to +5, a difference of 2, we should add 2 electrons to the right side of the oxidation half-reaction. AsO3^-3 --> AsO4^-3 + 2 e- The mass isn't balanced. There are 3 oxygen atoms on the left but 4 on the right. Since this reaction is occurring in an acidic environment, we'll add 1 water molecule to the left (to balance the oxygen atoms) and then 2 hydrogen ions on the right to compensate. AsO3^-3 + H2O --> AsO4^-3 + 2 H^+1 + 2 e- The oxidation half-reaction is complete. The reduction half-reaction will center around the nitrogen atom: NO3^-1 --> NO The nitrogen atom's O.N. drops from +5 to +2. Reduction involves a GAIN of electrons, and it'll take 3 electrons to cause the O.N. to fall three points. Add 3 electrons to the left-hand side of the half-reaction. NO3^-1 + 3 e- --> NO As before, the mass must be balanced with water and hydrogen ions: NO3^-1 + 4 H^+1 + 3 e- --> NO + 2 H2O The number of electrons released by the oxidation of arsenic must be the same as the number of electrons gained by the nitrogen atom. As the half-reactions sit now, they're NOT the same. The arsenic atom is losing 2 e- while the nitrogen atom gains 3 e-. In order to make them the same number, we'll have to multiply the half-reactions by some whole-number coefficients until the electron count balances. 3 x (AsO3^-3 + H2O --> AsO4^-3 + 2 H^+1 + 2 e-) 2 x (NO3^-1 + 4 H^+1 + 3 e- --> NO + 2 H2O) With the result being: 3 AsO3^-3 + 3 H2O --> 3 AsO4^-3 + 6 H^+1 + 6 e- 2 NO3^-1 + 8 H^+1 + 6 e- --> 2 NO + 4 H2O Now that the electron count is the same, we can recombine the half-reactions. 3 AsO3^-3 + 3 H2O + 2 NO3^-1 + 8 H^+1 + 6 e- --> 3 AsO4^-3 + 6 H^+1 + 6 e- + 2 NO + 4 H2O And cancel out or reduce anything that appears on both sides of the equation. 3 AsO3^-3 + 2 NO3^-1 + 2 H^+1 --> 3 AsO4^-3 + 2 NO + H2O Now you can bring the spectator ions back in: 3 H3AsO3 + 2 HNO3 --> 3 H3AsO4 + 2 NO + H2O And that does it. It's a good idea to check once again to ensure that the mass balances as it should. I hope that helps. Good luck!
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We found more questions related to the topic: How to write a mass balance equation chemistry

Originally Answered: How to balance this equation using REDOX?.?
This is a long process but will show you how. Each of these reactions is composed of two "half reactions", one for reduction where electrons are donated, and one for oxidation where electrons are gained. In each reaction after you get it balanced, electrons donated or lost by the reducing agent = electrons gained by the oxidizing agent. Start off by looking for elements that change valences or charge from the reactant side to the product side and place their TOTAL valence above them in the equation. In the first equation, inside the Cr2O7= ion, the two Cr atoms have a total charge of +12. and on the product side the one Cr has a charge of +3, so Cr undergoes electron gain going from a hlgher positive charge per atom to a lower positive charge, so it must be the OXIDIZING AGENT Write a half reaction for this and balace Cr atoms and show the number of electrons GAINED on the yields arrow. Assume the reaction takes place in an acid solution with available H+ to combine with the extra Oxygen atoms. 14 H+ + Cr2O7= yields 2Cr+3 + 7H2O above the yields arrow place Notice that the two Cr atoms on the left have a total charge of +l2 and the two Cr atoms on the right will have a total charge of +6, so a gain of 6electrons is noted on the yields arrow. The H+ ions available in acid solution will combine with the oxygen from the dichromate ion to make 7 H2O Now this half reaction is balanced. Next look for atoms that lose electrons and you will see that I- goes from negative one on the left to a positive 5 inside the IO3- ion. so write this half reaction as the I- is your reducing agent (electron donor) I- yields IO3- and note to go from a negative one charge to a plus 5 for a loss of 5 electrons, so place a 5e on the arrow. And add 3 Oxyget atoms to the left side to finish balancing the equation I- + 3O= yields IO3- Now multiply the number of electrons (on the arrow on the upper half reaction times the coefficients in the lower half reaction. And multiply the number of electrons on the lower half reaction times the coefficients in the upper half reaction Now add the two half reactions together and finish balancing up the H+ and H2O adding either to one side or the other to balance the equation. 34H+ + 5Cr2O7= + 6I- yields l0Cr+3 + 6IO3- + l7H2O On number 2 the two half reactions work out to be As Yields (loss of 3 electrons H3AsO3 and ClO3- yields (gain of 4 e) HClO It is not really necessary at this point to balance H+ and O, you can finish up doing that. Now multiply the two half reactions together as above and finish H and O balancing adding H+ and H2O as needed to balance to either side. Final result 3H+ plus HH2O + 4As plus 3ClO3- yields 4H3AsO3 plus 3HClO I hope this gets you started . in balancing redox equations.
Originally Answered: How to balance this equation using REDOX?.?
Here is my rules for balancing redox Rules for balancing Redox equations: 1) Determine the oxidation state (charge) of each atom of each element. 2) Determine which element’s atoms have changed oxidation number from the left side of the equation to the right side of the equation. 3) Write the equation as 2 half-reactions. Include the particles (atoms, ions, molecules) which are involved in change of oxidation state. 4) Balance all atoms and ions, except O and H, with the use of coefficients. 5) To balance the O’s, add enough water (H2O) to the side deficient in O atoms. 6) To balance the H atom, add enough H+ to the side deficient in H atoms. 7) Balance the charge by adding electrons to the side deficient in negative charges. 8) Multiply the half-reactions by smallest whole numbers to balance electrons. 9) Add the 2 half-reactions and subtract any duplications. If the reaction occurs in a basic solution, add the same number of OH-1ions to both sides to equalize the number of H+1 ions. Make me one of your contacts and send an example if you need more help!

Thutmosis
In a redox reaction, some of the atoms change their oxidation charge. You need to examine what has changed. For example MnO4- >>> Mn2+ In this half reaction, the Mn changes from 7+ to 2+, so it has gained 5 electrons.
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