Complete and balance the following half-reactions: A. Mo3+(aq)→Mo(s) (acidic solution) B. H2SO3(aq)→SO2−4(aq) (acidic solution) C. Complete and balance the following half-reacti… Show
Complete and balance the following half-reactions: A. Mo3+(aq)→Mo(s) (acidic solution) B. H2SO3(aq)→SO2−4(aq) (acidic solution) C. Complete and balance the following half-reaction: NO−3(aq)→NO(g)(acidic solution) D. CompleteComplete and balance the following half-reactions: A. Mo3+(aq)→Mo(s) (acidic solution) B. H2SO3(aq)→SO2−4(aq) (acidic solution) C. Complete and balance the following half-reaction: NO−3(aq)→NO(g)(acidic solution) D. CompleteBalancing redox reactions in acidic solutionFifteen ExamplesPoints to remember: 1) Electrons NEVER appear in a correct, final answer. In order to get the electrons in each half-reaction equal, one or both of the balanced half-reactions will be multiplied by a factor. After Example #5c, I have a suggestion for searching for help on balancing a specific equation. Example #1: ClO3¯ + SO2 ---> SO42¯ + Cl¯ Solution: 1) Split into unbalanced half-reactions: ClO3¯ ---> Cl¯ 2) Balance the half-reactions: 6e¯ + 6H+ + ClO3¯ ---> Cl¯ + 3H2O 3) Make the number of electrons equal: 6e¯ + 6H+ + ClO3¯ ---> Cl¯ + 3H2O 4) Add the two half-reactions for the final answer: ClO3¯ + 3H2O + 3SO2 ---> 3SO42¯ + Cl¯ + 6H+ Example #2a: H2S + NO3¯ ---> S8 + NO Solution: 1) The unbalanced half-reactions: H2S ---> S8 2) balance each half-reaction: 8H2S ---> S8 + 16H+ + 16e¯ 3) Make the number of electrons equal: 24H2S ---> 3S8 + 48H+ + 48e¯ <--- multiplied by a factor of 3 4) Add: 24H2S + 16H+ + 16NO3¯ ---> 3S8 + 16NO + 32H2O 5) Sometimes, you will see the nitric acid in molecular form: 24H2S + 16HNO3 ---> 3S8 + 16NO + 32H2O Example #2b: H2S + HNO3 ---> NO + S + H2O Discussion: Many times, teachers and textbooks will use S rather than S8. Notice that, in the answer, the S coefficient stays the same (but the subscript of 8 goes away) and the other coefficients are all reduced by a factor of 8. The answer to 2b is the exact same as 2a in terms of the stoichiometry of the reaction. Solution: 1) Half-reactions: NO3¯ ---> NO 2) Balance: 3e¯ + 4H+ + NO3¯ ---> NO + 2H2O 3) Equalize electrons: 6e¯ + 8H+ + 2NO3¯ ---> 2NO + 4H2O 4) Add: 3H2S + 2HNO3 ---> 2NO + 3S + 4H2O Example #3: MnO4¯ + H2S ---> Mn2+ + S8 Solution: 1) Half-reactions: H2S ---> S8 2) Balance: 8H2S ---> S8 + 16H+ + 16e¯ 3) Make the number of electrons equal (note that there are no common factors between 5 and 16 except 1): 40H2S ---> 5S8 + 80H+ + 80e¯ <--- factor of 5 4) The final answer: 40H2S + 48H+ + 16MnO4¯ ---> 5S8 + 16Mn2+ + 64H2O Example #4: Cu + SO42¯ ---> Cu2+ + SO2 Solution: 1) The unbalanced half-reactions: Cu ---> Cu2+ 2) The balanced half-reactions: Cu ---> Cu2+ + 2e¯ 3) The final answer: Cu + 4H+ + SO42¯ ---> Cu2+ + SO2 + 2H2O Example #5a: MnO4¯ + CH3OH ---> HCOOH + Mn2+ Solution: 1) The balanced half-reactions: 5e¯ + 8H+ + MnO4¯ ---> Mn2+ + 4H2O 2) Equalize electrons: 20e¯ + 32H+ + 4MnO4¯ ---> 4Mn2+ + 16H2O <--- factor of 4 3) The final answer: 12H+ + 5CH3OH + 4MnO4¯ ---> 5HCOOH + 4Mn2+ + 11H2O Example #5b: MnO4¯ + CH3OH ---> CH3COOH + Mn2+ Solution: 1) Half-reactions: MnO4¯ ---> Mn2+ 2) Balance: 5e¯ + 8H+ + MnO4¯ ---> Mn2+ + 4H2O 3) Equalize electrons: 20e¯ + 32H+ + 4MnO4¯ ---> 4Mn2+ + 16H2O 4) Add: 12H+ + 4MnO4¯ + 10CH3OH ---> 4Mn2+ + 5CH3COOH + 16H2O Example #5c: MnO4¯ + H2O2 ---> Mn2+ + O2 Solution: 1) Half reactions: MnO4¯ ---> Mn2+ 2) Balance: 5e¯ + 8H+ + MnO4¯ ---> Mn2+ + 4H2O 3) Equalize electrons: 10e¯ + 16H+ + 2MnO4¯ ---> 2Mn2+ + 8H2O 4) Add: 6H+ + 2MnO4¯ + 5H2O2 ---> 2Mn2+ + 5O2 + 8H2O The search suggestion is to type the reactants plus an arrow, like this: mno4- + h2o2 ---> Notice that I completely ignore capitals in the formulas. In the hits from that search, there will be a number of websites that will help you figure out the answer. However, be careful. Take a look at this website. You will see 14H2O rather than 8H2O. What happened? Answer: the writer of that page represented hydrogen ion as H3O+ rather than H+, thus adding six H2O to each side. Example #6: VO2+ + MnO4¯ ---> V(OH)4+ + Mn2+ Solution: 1) Half reactions: VO2+ ---> V(OH)4+ 2) Balance: 3H2O + VO2+ ---> V(OH)4+ + 2H+ + e¯ 3) Equalize electrons: 15H2O + 5VO2+ ---> 5V(OH)4+ + 10H+ + 5e¯ 4) Add: 11H2O + 5VO2+ + MnO4¯ ---> 5V(OH)4+ + Mn2+ + 2H+ Example #7: Cr2O72¯ + Cl¯ ---> Cr3+ + Cl2 Solution: 1) Half-reactions: Cr2O72¯ ---> Cr3+ 2) Balance: 6e¯ + 14H+ + Cr2O72¯ ---> 2Cr3+ + 7H2O 3) Equalize electrons: 6e¯ + 14H+ + Cr2O72¯ ---> 2Cr3+ + 7H2O 4) Add: 14H+ + Cr2O72¯ + 6Cl¯ ---> 2Cr3+ + 3Cl2 + 7H2O 5) A more detailed discussion about balancing this equation can be found here. 6) I once saw an unusual method to balancing this particular example equation. It winds up with the equation balanced in basic solution. Here it is, in all its glory: Cr2O72¯ + Cl¯ ---> Cr3+ + Cl2 + O2¯ 7) And then, since are in acidic solution, we use 14H+ to react with the hydroxide: Cr2O72¯ + 6Cl¯ + 7H2O + 14H+ ---> 2Cr3+ + 3Cl2 + 14H2O 8) And then remove seven waters from each side to arrive at the answer given in step 4. Example #8: MnO4¯ + S2¯ ---> MnS + S 1) Half-reactions: MnO4¯ ---> Mn2+ 2) Balance half-reactions: 5e¯ + 8H+ + MnO4¯ ---> Mn2+ + 4H2O 3) Equalize electrons: 10e¯ + 16H+ + 2MnO4¯ ---> 2Mn2+ + 8H2O 4) Add: 16H+ + 2MnO4¯ + 5S2¯ ---> 2Mn2+ + 5S + 8H2O 5) Add two sulfides on each side to make MnS: 16H+ + 2MnO4¯ + 7S2¯ ---> 2MnS + 5S + 8H2O 6) This document balances the equation in basic solution. This is an easy transformation from the answer in step 5, just add 16 hydroxides to each side: 8H2O + 2MnO4¯ + 7S2¯ ---> 2MnS + 5S + 16OH¯ 7) The linked document also keeps the MnS in the half-reaction and balances it with a sulfide on the left-hand side of the half-reaction. Example #9: As2S5(s) + NO3¯(aq) ---> H3AsO4(aq) + HSO4¯(aq) + NO2(g) Solution: 1) Half-reactions: 8H2O + As210+ ---> 2H3AsO4(aq) + 10H+ <--- note: neither reduction nor oxidation 2) Combine the first two half-reactions: 28H2O + As2S5 ---> 2H3AsO4(aq) + 5HSO4¯(aq) + 45H+ + 40e¯ 3) Add in the second half-reaction and equalize for electrons: 28H2O + As2S5 ---> 2H3AsO4(aq) + 5HSO4¯(aq) + 45H+ + 40e¯ 4) Add: 35H+ + As2S5 + 40NO3¯(aq) ---> 2H3AsO4(aq) + 5HSO4¯(aq) + 40NO2(g) + 12H2O Example #10: H3AsO3 + I2 ---> H3AsO4 + I¯ Solution: 1) Half-reactions: H3AsO3 ---> H3AsO4 2) Balance: H2O + H3AsO3 ---> H3AsO4 + 2H+ + 2e¯ 3) Electrons are already equal, so add: H2O + H3AsO3 + I2 ---> H3AsO4 + 2H+ + 2I¯ 4) If so needed, you could report this as fully molecular (instead of showing the HI - a strong acid - as fully ionized: H2O + H3AsO3 + I2 ---> H3AsO4 + 2HI Example #11: Balance the equation for the reaction of stannous ion with pertechnetate in acidic solution. Products are stannic ion, Sn4+ and technetium(IV), Tc4+ ions. Solution: 1) Net ionic: TcO4¯ + Sn2+ ---> Tc4+ + Sn4+ 2) Half-reactions: TcO4¯ ---> Tc4+ 3) Balance: 3e¯ + 8H+ + TcO4¯ ---> Tc4+ + 4H2O 4) Equalize electrons: 6e¯ + 16H+ + 2TcO4¯ ---> 2Tc4+ + 8H2O 5) Add: 16H+ + 2TcO4¯ + 3Sn2+ ---> 2Tc4+ + 3Sn4+ + 8H2O Example #12: H3AsO4 + Zn + HNO3 --> AsH3 + Zn(NO3)2 Solution: 1) Net-ionic form: AsO43¯ + Zn ---> AsH3 + Zn2+ 2) Half-reactions: AsO43¯ ---> AsH3 3) Balance: 8e¯ + 11H+ + AsO43¯ ---> AsH3 + 4H2O 4) Equalize electrons: 8e¯ + 8H+ + H3AsO4 ---> AsH3 + 4H2O <--- I used 3 H+ to recreate the H3AsO4 5) Add: 8H+ + H3AsO4 + 4Zn ---> AsH3 + 4Zn2+ + 4H2O 6) Add in eight nitrates: 8HNO3 + H3AsO4 + 4Zn ---> AsH3 + 4Zn(NO3)2 + 4H2O Example #13: O3 + Cl¯ ---> H2O + ClO3¯ Solution: 1) Half-reactions: O3 ---> H2O 2) Balance: 6e¯ + 6H+ + O3 ---> 3H2O 3) Electrons already equalized, so add: O3 + Cl¯ ---> ClO3¯ 4) Or, you can notice that dropping the water right at the start results in an equation balanced for atoms and for charge. Example #14: H2SO5 is named peroxymonosulfuric acid. One of its salts, KHSO5 (potassium peroxymonosulfate) is widely used as an oxidizing agent. Balance the following reaction in acidic solution: HSO5¯ + ClO2¯ ---> ClO3¯ + SO42¯ Solution: Comment: look to see if this one can be balanced for atoms and charge by sight. Hint: it can. The half-reaction method follows. 1) Separate into half-reactions: HSO5¯ ---> SO42¯ 2) Balance: 2e¯ + H+ + HSO5¯ ---> SO42¯ + H2O 3) Electrons already equal, so add: HSO5¯ + ClO2¯ ---> ClO3¯ + SO42¯ + H+ Example #15: As ---> H3AsO4 + AsH3 Solution: 1) Half-reactions: As ---> H3AsO4 2) Balance: 4H2O + As ---> H3AsO4 + 5H+ + 5e¯ 3) Equalize electrons: 12H2O + 3As ---> 3H3AsO4 + 15H+ + 15e¯ 4) Add: 12H2O + 8As ---> 3H3AsO4 + 5AsH3 Bonus Example: Cr2O72¯ + SO2 + H+ ---> Cr3+ + HSO4¯ + H2O Solution: 1) Half-reactions: Cr2O72¯ ---> Cr3+ 2) Balance in acidic solution: 6e¯ + 14H+ + Cr2O72¯ ---> 2Cr3+ + 7H2O 3) Equalize electrons: 6e¯ + 14H+ + Cr2O72¯ ---> 2Cr3+ + 7H2O 4) Add: 5H+ + Cr2O72¯ + 3SO2 ---> 2Cr3+ + 3HSO4¯ + H2O Sometimes you are given a net-ionic equation and asked to take it back to a full molecular equation. Sometimes, no context is added, so you have to make some informed predictions. Here's what I mean: Since the equation is in acidic solution, you can use HCl or HNO3. I'll use HCl. The most common dichromate that is soluble is potassium dichromate, so we will use that. Using those, we find this: 5HCl + K2Cr2O7 + 3SO2 ---> 2CrCl3 + 3KHSO4 + H2O However, there is a problem. One too many K and Cl on the right-hand side. The solution is to add one KCl to the left-hand side: KCl + 5HCl + K2Cr2O7 + 3SO2 ---> 2CrCl3 + 3KHSO4 + H2O You can write the equation using HNO3 and the nitrate would simply replace the chloride. Using sulfuric acid can be done but (and this is part of the informed prediction) probably should not. The chromium(III) ion is presented as an ion, meaning it's soluble. Chromium(III) sulfate is not soluble, which means you would have to write the full formula. Since that was not done, we conclude that the chromium ion was part of a soluble compound. This whole balance-a-redox-reaction-in-molecular-form is a thing and it's not covered very much in most textbooks. Here are some examples. How do you complete and balance a halfAcidic Conditions. Solution.. Step 1: Separate the half-reactions. ... . Step 2: Balance elements other than O and H. ... . Step 3: Add H2O to balance oxygen. ... . Step 4: Balance hydrogen by adding protons (H+). ... . Step 5: Balance the charge of each equation with electrons. ... . Step 6: Scale the reactions so that the electrons are equal.. How do you balance a halfIn general, the half-reactions are first balanced by atoms separately. Electrons are included in the half-reactions. These are then balanced so that the number of electrons lost is equal to the number of electrons gained. Finally, the two half-reactions are added back together.
What is the balanced chemical equation for the reaction in an acidic medium?Cr2O7(aq)^2 - + SO2(g)→ Cr(aq)^3 + + SO4(aq)^2 -
What is halfA half reaction (or half-cell reaction) is either the oxidation or reduction reaction component of a redox reaction. A half reaction is obtained by considering the change in oxidation states of individual substances involved in the redox reaction.
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