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Balancing Redox Equations - BR question

Discussion in 'MCAT Study Question Q&A' started by SSerenity, 05.12.14.

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  1. SSerenity

    SSerenity 2+ Year Member

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    [​IMG]



    I cannot for the life of me figure out who BR added those two half reactions.

    How do we go from having 12 OH- on the left side and 8OH- on the right side to having NO OH- on the left side and 2 OH- on the right side?

    We should have 4OH- on the LEFT side and NO OH- on the right side.

    We are balancing this in base btw.
     
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  3. SSerenity

    SSerenity 2+ Year Member

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    I missed the Q&A forum, can someone please move this?
     
  4. Czarcasm

    Czarcasm Hakuna matata, no worries. 2+ Year Member

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    I don't like TBR's approach to balancing equations. If you can give the original unbalanced equation, I could try to explain it in a way that's easier to remember.

    Basically what I do is I write the half reactions for each first. (I start with an acidic titration). Then I balance the actual atoms in each half reaction. Once atoms are balanced, I balance the overall charge by adding electrons on the appropriate side (goes to the side with more positive charge). You do this for both half reactions. Multiple each half reaction by least common multiple and add the half reactions together. The very last step you do is you balance any oxygens by adding water to the appropriate side (to make it even on both sides). Then you balance the extra protons with H+ (as if you're doing an acidic titration). So if for example, the reactants have 8 H2O, add 16 H+ to products.

    For basic titrations you have to go one extra step and neutralize the acid (H+) with OH-. What you do to one side you must do to the other. So, in my example of having 16 H+ in products, you add 16 OH- to both reactants and products. 16 OH- stays on reactants, but the neutralizing acid and base yields 16 H2O in products. Note that in my example, I already had 8 H2O in reactants. So overall, 8 of those will cancel out with products and leave a net of 8 H2O in products remaining.
     
  5. SSerenity

    SSerenity 2+ Year Member

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    [​IMG]

    I tried it your way, and I get the same two balanced redox equations.
    However, when I add them I get 4 Hydroxides on the left and none on the right.

    Did BR just totally screw up their addition here? I think the accidentally used Zn(OH)2 in their final calculations instead.
     
    Last edited: 05.12.14
  6. Czarcasm

    Czarcasm Hakuna matata, no worries. 2+ Year Member

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    Two half-reactions:

    Reduction: 3e + MnO4-(aq) --> MnO2 (s) (OS on Mn goes from: +7 to +4) : x2
    Oxidation: Zn(s) --> Zn(OH)4^2-(aq) + 2e (OS on Zn goes from: O to +2 : x3
    ---

    6e + 2MnO4- --> 2MnO2
    3Zn --> 3Zn(OH)4^2- + 6e
    ---------------------------------
    3Zn + 2MnO4- ---> 2MnO2 + 3Zn(OH)4^2-

    Balance the oxygens: Reactants have 8 oxygens; Products have 12. Therefore, add 4 H2O to reactant side:
    3Zn + 2MnO4 + 4H2O ----> 2MnO2 + 3Zn(OH)4^2-

    Now balance hydrogens with H+: 8H on reactant side, 12 on product side. Add 4H+ to reactant side:
    3Zn + 2MnO4 + 4H2O + 4H+ ----> 2MnO2 + 3Zn(OH)4^2-

    Neutralize all H+ by adding OH- to both sides. In this case, we add 4OH- to both sides to neutralize H+ on reactants to more H2O.
    3Zn + 2MnO4 + 4H2O + 4H2O ----> 2MnO2 + 3Zn(OH)4^2- + 4OH-

    And the final balanced redox reaction should come out to:
    3Zn + 2MnO4 + 8H2O ----> 2MnO2 + 3Zn(OH)4^2- + 4OH-

    Edit: I glanced again at their balanced reaction and the hydrogens & oxygens don't balance out on both sides, so they made a mistake somewhere. Should of been the first red flag, but I didn't bother reading their explanation/answer, lol.
     
  7. SSerenity

    SSerenity 2+ Year Member

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    Thanks for the explanation!

    Ok good. That question was driving me mad lol.
     
    Czarcasm likes this.
  8. BerkReviewTeach

    BerkReviewTeach Company Rep & Bad Singer Exhibitor 7+ Year Member

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    Yeah, that was a typo in an older edition of the general chemistry book.

    Sorry to be the bringer of bad news, but if you look at your final balanced equation, it's not quite right. You show 3 Zn, 2 Mn, 16 O, 16 H, and a -2 charge on the left side and 3 Zn, 2 Mn, 20 O, 16 H, and a -10 charge on the right side.

    The balance equation should be:
    3Zn + 2MnO4^- + 4OH^- + 4H2O ----> 2MnO2 + 3Zn(OH)4^2-

    But truth be told, when it comes to balancing questions, you should look for the reductant/oxidant ratio and then look at the answer choices with the correct ratio and eliminate the ones that are not correctly balanced (either wrong number of atoms on each side or an imbalance of charges on each side). Balancing from scratch would not be the best use of a test taker's time.
     
  9. Czarcasm

    Czarcasm Hakuna matata, no worries. 2+ Year Member

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    Nice catch. Darnit though, lol! I swear I counted it twice and still miscalculated -sigh-. In this scenario, using my method, you'd have to keep going a few extra steps to get the O's to balance (usually it balances out after the first step of adding OH- but in few scenarios like this, it doesn't ...and so you must keep balancing by adding: H2O for extra O's, then H+ for extra H's, then OH- to neutralize H+'s). Quoted:

    Like you suggested, for a test like the MCAT, it's unlikely they'd ask for a full balanced reaction like this I think. Usually it'll be something simple like two metals and their ions. But if by some chance they did, personally I would just calculate the charges and check to see if the atoms balance for each answer choice (provided I don't miscalculate lol). I think that'd be a much faster approach though.
     
    Last edited: 05.13.14
    SSerenity likes this.

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