Hi, I\'m having trouble understanding how to do the followingproblem: Write bala
ID: 680942 • Letter: H
Question
Hi,I'm having trouble understanding how to do the followingproblem:
Write balanced net ionic equations for the followingoxidation-reduction reactions that occur in acidic solution.
a) a solution of KI is treated with HNO3 (assumingNO3- is reduced to NO)
b) Potassium iodide is added to an unknown containingNO2- (assuming reduction to NO)
c)Potassium permanganate is added to a solution of KI (assumingreduction to Mn2+)
d) Potassium permanganate is added to a solution ofNa2S; a yellowish white solid forms.
If you could please explain how to solve this problem by showingall of the steps, I would really appreciate it.
Explanation / Answer
For oxidation-reduction (redox) problems, one of the substances isbeing oxidized, the other is being reduced. In order to tell whichis which, the substance oxidized loses electrons, and so it becomesmore positive, while the substance that is reduced gains electronsand so becomes more negative. The best way to begin a redox equation is by first finding the waythe reaction will occur. That is, setting up an equation. Forthese, the equation will be unbalanced to start with. So, for A) A solution of KI is treated with HNO3 it isnecessary to know what each of these will form. Sometimes it isdifficult to tell what the reaction will form, but if you havetrouble, you can find a standard reduction potential table ineither your chemistry book or online that you can use to help youfigure out how the reaction occurs. Already, you know that NO3- will be changedinto NO. This helps you figure out what KI (which will beI- in your equation) will change into. The easiest wayto do this is to sign oxidation numbers to the elements in eachsubstance: +5 -2 +2 -2 NO3- --> NO These should add up to the overall charge of the substance. ((ThreeO with -2= -6 charge) + (One N with +5= +5 charge) = -1 charge onNO3- ) While oxygen keepsits same oxidation number throughout the reaction, nitrogen goesfrom +5 to +2. It is becoming less positive, therefore it isgaining electrons, so it is reduced in this reaction. Thisindicates that the KI will be reduced. So the reaction for a would be: I- + NO3- --> NO + I2 But this is unbalanced. In order to balance thereaction, it is necessary to split it up into its two halfreactions: I- -->I2 NO3- --> NO Next, you have to balance each half reaction in a series ofsteps: 1. Balance all elements other than oxygen and hydrogen. 2. Balance oxygen by adding H2O to the side with lessoxygen. 3. Balance hydrogen by adding H+ to the side with lesshydrogen. 4. Balance charge, by adding electrons to the side with a morepositive total charge. 5. Multiply half reactions so that they have the same number ofelectrons on opposite sides (If your electrons are not on oppositesides from another, then something is wrong.) 6. Add the half reactions together. 7. Cancel out any substances on both sides of the reaction. 8. If it is in acidic solution, do not do this step. If it is basicsolution, add OH- to balance out H+ and toform water. To demonstrate this: I- -->I2 NO3- --> NO 1. 2I- -->I2 1. NO3- --> NO 2, 3 are notapplicable 2. NO3- --> NO +2H2O 4. 2I- --> I2 +2e- (e- denoteselectrons) 3. 4H+ + NO3- -->NO + 2H2O 4. 3e- + 4H+ + NO3- -->NO + 2H2O 5. 3(2I- --> I2 +2e-) 2( 3e- + 4H+ + NO3---> NO + 2H2O) 6I- --> 3I2 +6e- 6e- + 8H+ + 2NO3---> 2NO + 4H2O 6. 6I- + 6e- + 8H+ +2NO3- --> 2NO + 4H2O+3I2 + 6e- The 6e- cancel. For all redox equations, electronsshould always cancel. If they do not, it is indicative that thereaction is improperly balanced. Some other redox reactions willhave other substances that balance, too. 7. 6I- + 8H+ + 2NO3---> 2NO + 4H2O +3I2 The others work exactly the same way. The extra information givento you will help you determine what each substance willdo.
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