Assignment 90 BALANCING OF REDOX EQUATIONS

Transcription

1 Assignment 90 BALANCING OF REDOX EQUATIONS This Assignment concerns balancing of chemical equations for "redox reactions." Balancing of this kind must be done in a stepwise manner. The Procedures given below are a specific method which we recommend -- sometimes referred to as the "Ion - Electron Method". In this course ALL students are expected to become familiar with, and to demonstrate, THIS METHOD of balancing for work concerned with redox reaction equations. NOTE: In all problems of this Assignment, all the major reactants and products of a reaction will be specified, as given skeletal data, EXCEPT FOR common aqueous species such as H 2 O, H +, and OH. (For the purposes of redox balancing, the formula, H +, is adopted as the symbol for hydronium ion (instead of the usual H 3 O + ), so as to ease the balancing procedure.) It is assumed that you already can accurately assign Ox.Nos., as described in the F&R manual. How to Balance Redox Rxn Equations Redox refers to a type of reaction -- Reduction and Oxidation. Redox reactions involve transfer of electrons between reactants. When this happens there is always a change in Ox.No. Reactant Product LOW Ox.No. HIGH Ox.No. Oxidation is occurring HIGH Ox.No. LOW Ox.No. Reduction is occurring You always have both. You need to identify by formula the reactants and products with respect to "what has been oxidized" and "what has been reduced" -- the Befores in each case and the Afters in each case. This identification, by formula, of the Befores and the Afters Pair (in terms of Ox.No.) is known as Specifying the Redox Couple. The formulas will always be present in the initial skeletal reaction, but you need to identify them. The Procedure: STAGE I. 1. Find the two redox couples within the skeletal reaction. (Use the formulas exactly as given.) STAGE II. Take each redox couple in turn through steps 2, 3, 4, 5: 2. Balance the element(s) that change Ox.No. (Do this by placing coefficients in front of the formulas, but be careful not to alter any given formula.) 3. Then balance the atoms of the other elements -- If other than O or H, just add the needed element(s) as spectator ions. For balancing O, use H 2 O where needed so as to complete the oxygen balance. Then, for balancing H, use H + where needed to complete the atom balance. (See below for examples of the technique. Hydrogen is the last element to get balanced.)

2 STAGE II (continued) 4. Now, charge balance with e 's (so that each side of equation has same net charge). 5. CHECK the initial acid/base conditions (these will be specified with the reaction). If acidic -- what you have balanced is O.K. as is. If alkaline (basic) --"neutralize" any H + by adding to BOTH sides of equation the equivalent # of OH 's (also clean up the resulting H 2 O 's). If neutral -- make sure the left-hand sides of equations have only H 2 O 's. (No H + nor OH is permitted on the left-hand side -- they are O.K. on the right-hand side.) Do what needs doing. STAGE III. At this point in the Procedure you should have two balanced equations (each is called a "halfreaction" equation). In one of them, the e 's are a product of the rxn. In the other one the e! 's occur as a reactant. These will now be combined into a single net equation (by steps 6 and 7): 6. Now balance both half-rxns together (eliminating the e 's) so that the final reaction shows no electrons remaining free. (Each half-rxn may need to be multiplied all the way through by a factor, such as 2 or 3, so that the e 's can cancel out in the overall net equation.) In the complete equation there must be NO e 's. 7. Finally, check for any redundancies, so that the final (combined) eqn is in simplest form. Note: We are providing, as given information, not only the formulas of the main reactants, but also the formulas of the main products as well, because predicting the principal products of a redox reaction under specified conditions is an ability gained only through much experience in working with redox reactions. Since many elements can have more than two oxidation states, it may be difficult without experimental evidence to decide which of several possible products should be expected. The only missing formulas that we may expect you to supply during the balancing procedure are obvious ones such as H 2 O and H + and OH. Do NOT add O 2 or H 2 gases for purposes of balancing (if needed, they would be specified among the given substances). Also, the O 2 anion is NEVER used because it cannot exist in aqueous solutions. A special advantage to be gained from this particular method of balancing redox reactions is the deducing of the "half-reactions", which themselves are important in later course work (see note***). ***In some high school courses and some textbooks, other methods are described (such as the use of oxidation numbers) which are less generally applicable; those methods will not be used in the present course. The "ion-electron method" described in this Assignment is applicable for all aqueous redox reactions as well as for electrical cells, and we expect all students to become adept in using it. Examples of balancing a "half-reaction" The examples below emphasize the thought-process of STAGE II (steps 2, 3, 4, 5) in more detailed fashion

3 Example A: Suppose a redox reaction in acid solution involves a redox couple, MnO 4 and Mn 2+, undergoing a reduction half-reaction. How to balance the half-reaction? Step 1: To be balanced: MnO 4 acid Mn 2+ The Mn atoms are changing Ox.No. (from ox.no.+7 on the Befores side to ox.no.+2 on the Afters side). Step 2: The Mn atoms are already balanced in the skeletal equation as originally written (coefficients of 1 being implied) -- one Mn atom on left side and one Mn atom on right side, as originally written. So, proceed to balance the "other" elements by Step 3. Step 3: Four {O} atoms are already present on left side (i.e., in permanganate ion); none are on the right. Use H 2 O molecules where needed (i.e., four of them on the right), so as to provide the four {O} atoms needed for balance. So do this: MnO 4 Mn H 2 O Oxygen atoms now balance, but an imbalance has appeared in hydrogen. So as to balance the 8 {H} already present on the right, use H + ions where needed (i.e., eight of them on the left), so as to provide the eight {H} needed for balance. So do this: MnO H + Mn H 2 O Step 4: All the atoms now balance, but the electrostatic charges have not yet been balanced. There is a total charge of 7+ on left and a total charge of 2+ on right. Balance the charges by adding electrons (e ) to the equation -- five electrons are needed on the left side to equalize the charges. So do this: 5 e + MnO H + Mn H 2 O Step 5: With atoms and charge all balance, now CHECK the given condition for the reaction. Since the given condition ("acid" solution) agrees with the presence of H + in the equation, the final "half-reaction equation" is O.K. as written. acid Example B: Balance this given half-reaction: HNO 2 NH 4 + Steps 1 & 2: The element nitrogen has the differing Ox.Nos., and the N atoms are already balanced. Step 3: No "other" elements need balancing except {O} and {H} atoms. Balance oxygen first. Two {O} atoms are already present in HNO 2 on the left; none on the right. Use H 2 O molecules where needed to achieve oxygen balance (i.e., two of them on the right): HNO 2 NH H 2 O With oxygen balanced, now 8 {H} atoms are present on the right; whereas 1 {H} atom is present on the left. Use H + ions where needed to achieve hydrogen balance (i.e., seven of them on the left): HNO H + NH H 2 O All atoms are now in balanced condition. Step 4: There is yet imbalance in total electrostatic charge (7+ on left side and 1+ on right side). Balance the charges by adding electrons to the equation where needed (i.e., 6 electrons on the left): 6 e + HNO H + NH H 2 O Now there is a total charge of 1+ on each side

4 Step 5: CHECK the given condition for the reaction. Since the given "acid" condition agrees with the presence of H + in the equation, the "half-reaction" equation is O.K. as written. Example C: Balance this given half-reaction: NO 2 base NH 3 Steps 1 & 2: The element nitrogen has the differing Ox.Nos., and the N atoms are already balanced. Step 3: No "other" elements need balancing except {O} and {H} atoms. Balance oxygen first. Two {O} are already present within NO 2 on left; none on right. Use H 2 O molecules to balance. NO 2 NH H 2 O Also, a total of 7 {H} on right need to be balanced. Use H + ions to balance them. NO H + NH H 2 O All atoms now balance, but the electrostatic charges do not. Step 4: Balance the electrostatic charges ( 7+ on left and 0 on right ) by means of electrons. 7 e + NO H + NH H 2 O Step 5: CHECK the given condition, "base". To accommodate this condition, you must "neutralize away" any H + in the equation -- but DON'T UNBALANCE the equation! Therefore, add appropriate amounts of OH to both sides of the equation -- a total of 7 OH to each side. On the left side, the effect of the added 7 OH is to convert each H + into a H 2 O molecule. 7 e + NO H 2 O NH H 2 O + 7 OH Atoms and charges still balance, because you added the same amount ( 7 OH ) to each side. Now cancel the two redundant H 2 O molecules from both sides. (Note that the equation REMAINS BALANCED so long as you do the same action -- removal of 2 H 2 O -- on both sides.) Net result (properly balanced): 7 e + NO H 2 O NH OH Example D: Given the half-reaction: CNO neutral HCN 1 {O} on left needs balancing, so do this: CNO HCN + H 2 O 3 {H} on right needs balancing, so do this: CNO + 3 H + HCN + H 2 O (Atoms now balance.) Balance the charges: 2 e + CNO + 3 H + HCN + H 2 O (Atoms and charges now balance.) For "neutral" conditions, it is not appropriate to keep H + as a reactant (on the left), so the 3 H + must be "neutralized away" by adding 3 OH to both sides -- maintaining the balance: 2 e + CNO + 3 H 2 O HCN + H 2 O + 3 OH Then clean up the redundancy. Net "half-reaction": 2 e + CNO + 2 H 2 O HCN + 3 OH (balanced)

5 neutral Example E: Given the half-reaction: HCN CNO Note: Although this involves the same redox couple as Example D, here oxidation is occurring rather than reduction, and the neutral "half-reaction" comes out quite differently. 1 {O} on right needs balancing, so do this: HCN + H 2 O CNO 3 {H} on left needs balancing, so do this: HCN + H 2 O CNO + 3 H + (The atoms now balance.) Balance the charges: HCN + H 2 O CNO + 3 H e For "neutral" conditions, this half-reaction is O.K. as is. Water is O.K. as a reactant, and H + is O.K. when it is a product (on the right side). Examples of balancing a complete reaction Note: The reactions shown below are for illustration only -- some may not actually proceed as written here. In given equations, it is customary to indicate a gas by means of a parenthetical letter, (g), and a solid by means of an (s). These notations need to be included in the final balanced equation (only). 1) Balance the reaction: NO 3 + H 2 S acid NO(g) + S 8 (s) Half-reactions include: i) NO 3 NO (Reduction occurring; nitrogen ox.nos. +5 to +2.) ii) H 2 S S 8 (Oxidation occurring; sulfur ox.nos. 2 to 0.) To balance the atoms in each half-reaction, use only H 2 O and H + as additional species: i) NO 3 NO showing 2 {O} on left needing to be balanced. NO 3 NO + 2 H 2 O showing 4 {H} on right needing to be balanced. NO H + NO + 2 H 2 O showing charges yet unbalanced (+3 on left and 0 on right). 3 e + NO H + NO + 2 H 2 O (The nuclei and electrons balance now.) ii) H 2 S S 8 showing sulfur needing to be balanced. 8 H 2 S S 8 showing 16 {H} on left needing to be balanced. 8 H 2 S S H + showing charges yet unbalanced (0 on left and +16 on right). 8 H 2 S S H e (The nuclei and electrons balance now.) The two balanced "half-reactions" need to be combined into a single "redox equation" by Steps 6 and 7. To equate electrons, (i) must accept 16 3 electrons, and (ii) must donate 3 16 electrons; Therefore, multiply (i) 16 and (ii) 3, then add (cancelling the electrons): 48 e + 16 NO H + 16 NO + 32 H 2 O 24 H 2 S 3 S H e

6 16 NO H H 2 S 16 NO + 32 H 2 O + 3 S H + Cancel 48 H + on both sides (superfluous quantities) and complete the phase notations for gas and for solid: 16 NO H H 2 S 16 NO(g) + 32 H 2 O + 3 S 8 (s) The final equation shows a balancing of electrical charge as well as the elements, and the coefficients are in the smallest possible whole numbers. It also is suitable for "acid" conditions. 2) Balance the reaction: Cr 2 O SO 2 (aq) acid Cr 3+ + HSO 4 Half-reactions involve: 2 i) Cr 2 O 7 Cr 3+ (Reduction of chromium from ox.nos. +6 to +3) ii) SO 2 HSO 4 (Oxidation of sulfur from ox.nos. +4 to +6) To balance each half-reaction, use only H 2 O and H + : 2 i) Cr 2 O 7 2 Cr 3+ (note two Cr 3+ ions; 7 {O} on left needs balancing) 2 Cr 2 O 7 2 Cr H 2 O ( 14 {H} on right needs to be balanced. ) 2 Cr 2 O H + 2 Cr H 2 O (charges yet unbalanced; 12+ on left, 6+ on right) 6 e 2 + Cr 2 O H + 2 Cr H 2 O (half-reaction is now balanced) ii) SO 2 HSO 4 SO H 2 O HSO 4 SO H 2 O HSO 4 SO H 2 O HSO 4 (2 {O} on right needs balancing) (3 {H} on left now needs balancing) + 3 H + (charges yet unbalanced;: 0 on left, 2+ on right) + 3 H e (half-reaction is now balanced) To equate electrons, multiply equation (ii) by 3 and (i) by 1; and then add: 6 e 2 + Cr 2 O H + 2 Cr H 2 O 3 SO H 2 O 3 HSO H e 2 Cr 2 O H SO H 2 O 2 Cr H 2 O + 3 HSO 4 Cancel H 2 O and H + redundancies: Cr 2 O H SO 2 (aq) 2 Cr 3+ + H 2 O + 3 HSO H + (balanced "redox eqn") Always CHECK your final equation to ensure that the atoms and the charges are indeed balancing, and the coefficients are in the simplest possible ratio. 3) Balance: NO 3 + Zn(s) base NH 3 (g) + ZnO 2 2 Half-reactions involve: 2 i) Zn ZnO 2 (Oxidation of zinc from ox.nos. 0 to +2) ii) NO 3 NH 3 (Reduction of nitrogen from ox.nos. +5 to 3) To balance each half-reaction, start by using only H 2 O and H + as additional species:

7 i) Zn 2 ZnO 2 (Zn is balanced; but 2 {O} on right needs balancing) Zn + 2 H 2 O 2 ZnO 2 ( 4 {H} on left needs to be balanced yet.) Zn + 2 H 2 O 2 ZnO H + (charges yet unbalanced: 0 on left, 2+ on right) Zn + 2 H 2 O 2 ZnO H e (balanced; but H + must be "neutralized" away) Zn + 2 H 2 O + 4 OH 2 ZnO H 2 O + 2 e (yet needs to be simplified) Zn + 4 OH 2 ZnO H 2 O + 2 e (half-rxn is now balanced and appropriate) ii) NO 3 NH 3 (N is balanced; 3 {O} on left needs to be balanced) NO 3 NH H 2 O (the 9 {H} on right needs to be balanced) NO H + NH H 2 O (charges yet unbalanced; 8+ on left, 0 on right) 8 e + NO H + NH H 2 O (balanced; but H + must be "neutralized" away) 8 e + NO H 2 O NH H 2 O + 9 OH (yet needs to be simplified) 8 e + NO H 2 O NH OH (half-rxn now balanced and appropriate) To equate electrons, 4 (i) and 1 (ii): 4 Zn + 16 OH 2 4 ZnO H 2 O + 8 e 8 e + NO H 2 O NH OH 4 Zn + 16 OH + NO H 2 O 4 ZnO H 2 O + NH OH Cancel H 2 O and OH, and complete the phase notations: 4 Zn(s) + 7 OH 2 + NO 3 4 ZnO H 2 O + NH 3 (g) (balanced "redox" eqn) 4) Balance: As 2 S 5 (s) + NO 3 acid Half-reactions involve: i) As 2 S 5 H 3 AsO 4 + HSO 4 ii) NO 3 NO 2 To balance each, use only H 2 O and H + : NO 2 (g) + H 3 AsO 4 + HSO 4 both products are needed, so as to include both the arsenic and sulfur elements i) As 2 S 5 2 H 3 AsO HSO 4 (to balance the As and S atoms) But 28 {O} on right need balancing yet. Use H 2 O for balancing: As 2 S H 2 O 2 H 3 AsO HSO 4 There is a total of 56 {H} on left and a total of 11 {H} on right, so a net 45 {H} on left needs balancing. Use H + for balancing: As 2 S H 2 O 2 H 3 AsO HSO H + There are total charges of 0 on left, +40 on right. Use electrons to balance: As 2 S H 2 O 2 H 3 AsO HSO H e (This half-reaction is now balanced and is appropriate for "acid" conditions.)

8 ii) NO 3 NO 2 NO 3 NO 2 + H 2 O NO H + NO 2 + H 2 O e + NO H + NO 2 + H 2 O (Half-rxn is balanced and O.K. for"acid"conditions) Combine the half-rxns into a "redox equation." To equate electrons, 40 Η (ii) and 1 Η (i): As 2 S H 2 O 2 H 3 AsO HSO H e 40 e + 40 NO H + 40 NO H 2 O As 2 S 5 (s) + 40 NO H + 2 H 3 AsO HSO NO 2 (g) + 12 H 2 O (with excess H 2 O and H + cancelled, and solid and gas denoted) 5) Balance: NO(g) + BiOOH(s) Half-reactions involve: i) NO HNO 2 ii) BiOOH Bi To balance, start by adding H 2 O and H + : neutral Bi(s) + HNO 2 i) NO HNO 2 ( N is already balanced. 1 {O} on right needs balancing ) NO + H 2 O HNO 2 ( 1 {H} on left needs balancing ) NO + H 2 O HNO 2 + H + ( with elements balanced, charges need balancing) NO + H 2 O HNO 2 + H + + e (Half-rxn is balanced and O.K. for "neutral" ) (Note: having H 2 O on left is O.K. for "neutral" conditions.) ii) BiOOH Bi ( Bi is already balanced. 2 {O} on left need balancing ) BiOOH Bi + 2 H 2 O ( 3 {H} on right need balancing ) BiOOH + 3 H + Bi + 2 H 2 O (elements balanced; charges need balancing) 3 e + BiOOH + 3 H + Bi + 2 H 2 O (Half-rxn has elements and electrons balanced; but having H + on left is NOT O.K. for "neutral". ) To accommodate the given "neutral" conditions, add 3 OH to both sides, so as to "neutralize" away the 3 H + but not to disrupt the balancing: 3 e + BiOOH + 3 H 2 O Bi + 2 H 2 O + 3 OH Now cancel the redundant H 2 O: 3 e + BiOOH + H 2 O Bi + 3 OH (properly balanced half-reaction) Combine the two appropriate half-reactions into a single "redox" equation. To equate electrons, 3 (i) and 1 (ii): 3 NO + 3 H 2 O 3 HNO H e 3 e + BiOOH + H 2 O Bi + 3 OH 3 NO + BiOOH + 4 H 2 O 3 HNO 2 + Bi + 3 H OH 3 HNO 2 + Bi + 3 H 2 O

9 Finally, cancel the excess H 2 O, and denote the gas and solid states: 3 NO(g) + BiOOH(s) + H 2 O 3 HNO 2 + Bi(s) (balanced "redox eqn") Always watch for the possibility of simplifying the final equation. Note in the above example, the H + and OH products MUST be combined into H 2 O, and the excess H 2 O then MUST be cancelled to give the simplest equation. Sometimes all the coefficients in the complete equation can be divided by a number so as to obtain a simpler whole number ratio -- as in the following example. 6) Balance: (CN) 2 base CNO + CN Half-reactions involved: i) (CN) 2 CNO ii) (CN) 2 CN (Note that the same substance is being used as a reactant in both half-reactions.) To balance each half-reaction: i) (CN) 2 2 CNO ( to balance C and N ) (CN) H 2 O 2 CNO ( to balance the 2 {O} in usual manner ) (CN) H 2 O 2 CNO + 4 H + ( to balance the 4 {H} in usual manner ) (CN) H 2 O 2 CNO + 4 H e ( now atom-balanced and charge-balanced ) The half-rxn does not yet conform to the given "base" conditions, however. Any H + must be "neutralized away," regardless where found in the equation. But, in order to maintain the equation's balanced state during Step 5, any further actions must be done to both sides of equation. So, 4 OH must be added to each side: (CN) H 2 O + 4 OH 2 CNO + 4 H 2 O + 2 e ( to "neutralize" the 4 H + ) (CN) OH 2 CNO + 2 H 2 O + 2 e ( to eliminate redundancy ) This half-reaction remains balanced but is now O.K. for the stated "base" conditions. ii) (CN) 2 2 CN ( to balance C and N ) 2 e + (CN) 2 2 CN ( to balance it electrostatically ) This half-reaction is balanced and also O.K. for "base" conditions. To equate electrons, combine (i) and (ii). The result: 2 (CN) OH 2 CNO + 2 H 2 O + 2 CN Simplify, by dividing all coefficients by 2: (CN) OH CNO + H 2 O + CN This "redox eqn" is now in appropriately balanced form. Note: The "Rules of the Game" REQUIRE that every formula be written correctly (the proper electrostatic charge included). Any error in writing of any original formula will yield a false equation, regardless how carefully the balancing is performed!!!

10 Recognizing changes of Ox.No. in reactions NOTE: This brief topic is merely a Reiteration / Review of earlier material already dealt with in the course. In the earlier Formulas & Reactions manual, emphasis was placed on oxidation numbers (ox.no.) and the rules for assigning an ox.no. to each element in a formula of a species (molecular or ionic). Oxidation numbers are used to recognize redox reactions, and they often are used in names of chemical substances. Redox reactions (electron transfer reactions) are those having changes in ox.no. In reactions involving only simple ions, the ox.no. changes are revealed by the changes in the ionic charges (recall that the ox.no. of a simple ion equals the charge of that ion). For example: Fe 2+ + Ce 4+ Fe 3+ + Ce 3+ The ox.no. of Fe changes from +2 to +3, and the ox.no. of Ce from +4 to +3. One species (Fe 2+ ) is being oxidized, an increase in ox.no., while another species (Ce 4+ ) is being reduced, a decrease in ox.no. For redox reactions involving complex species, the changes in ox.no. are less obvious, but the changes may be found by assigning ox.no.'s. to each element according to the usual ox.no. rules. For example: 2 Mn BiO 3 reactants + 14 H + 2 MnO Bi H 2 O products H ox.no. +1 ; H ox.no. +1 ; O ox.no. 2 ; O ox.no. 2 ; Bi ox.no. +5, Bi ox.no. +3 ; since Bi + [3 x (2)] = 1 Mn ox.no. +7, and thus Bi = 1 [3 x (2)] ; since Mn + [4 x (2)] = 1 Mn ox.no. +2 and thus Mn = 1 [4 x (2)] Mn is being oxidized (+2 to +7), an increase of 5 units; while Bi is being reduced (+5 to +3), a decrease of 2 units. Note that we do not need a complete balanced equation in order to determine the changes in ox.no. values. What is necessary is the information that Mn 2+ as a reactant ends up as MnO 4 as a product and that BiO 3 as a reactant ends up as Bi 3+ as a product. More practice (see the five transformations below): Analyze the ox.no. changes occurring in these transformations:

11 1) Cl Cl 2 Cl ox.no. 1 Cl ox.no. 0 Cl changes from 1 to 0; increases one unit in ox.no.; is (being) oxidized. 2) CrO 4 2 Cr 2 O 7 2 O ox.no. 2 ; O ox.no. 2 ; Cr ox.no. +6, Cr ox.no. +6, since Cr + [4 (2)] = 2 since (2 Cr) + [7 (2)] = 2 and thus Cr = 2! [4 (2)] whereby (2 Cr) = 2 [7 (2)] and thus Cr = { 2 [7 (2)] } / 2 No change in any ox.no.; no oxidation or reduction has occurred -- not a redox reaction. 3) NO 3 NH 4 + O ox.no. 2 ; H ox.no. +1 ; N ox.no. +5 N ox.no. 3 N changes from +5 to 3; decreases 8 units in ox.no.; is (being) reduced. 4) Hg 2 Cl 2 (s) HgCl 2 Cl ox.no. 1 ; Cl ox.no. 1 ; Hg ox.no. +1, Hg ox.no. +2, since (2 Hg) + [2 Η (1)] = 0 since Hg + [2 (1)] = 0 and thus (2 Hg) = +2 Hg changes from +1 to +2; increases 1 unit in ox.no.; is (being) oxidized. 5) Al(OH) 3 (s) Al(OH) 4 H ox.no. + 1 O ox.no.- 2 or (OH) ox.no. 1 H ox.no. + 1 O ox.no.- 2 or (OH) ox.no. 1 Al ox.no. +3, Al ox.no. +3, since Al + [3 (+12)] = 0 since Al + [4 (+12)] = 1 which is Al + [3 (1)] = 0 which is Al + [4 (1)] = 1 No change in any ox.no.; no oxidation nor reduction

12 A way to CHECK for correct balancing of half-reactions Ox.nos. provide an independent means for checking whether a half-reaction has been balanced correctly. The number of electrons appearing in a balanced half-reaction should equal the total change in ox.no. (i.e., the sum of all ox.no. changes for all atoms undergoing ox.no. change in the half-reaction). Thus, it is possible to check whether a half-reaction has been balanced correctly, by analyzing the ox.no.'s. of the element(s) being oxidized or reduced. Examples: Use ox.no. calculations to CHECK whether these given half-reactions are balanced. 1) As 2 O H 2 O 2 H 3 AsO H e The element As changes ox.no. from +3 to +5; oxidation (= loss of electrons); change of 2 units (per As). Since two As atoms are included in the half-reaction, the total change in ox.no. is 2 (2) = 4 units, which agrees with the four electrons shown in the balanced half-reaction. 2) 5 e + MnO H + Mn H 2 O Mn changes ox.no. from +7 to +2; reduction (= gaining of electrons); change of 5 units (per Mn). Since one Mn atom is involved in the reaction, the total change in ox.no. is 1 5 = 5 units, which agrees with the five electrons shown in the balanced half-reaction. 3) 2 e + (CN) 2 2 CN The ox.no. of the atom-pair, (CN), may be considered to change from 0 to 1; reduction (= gain of electrons); a change of 1 unit per (CN). Since two (CN) atom-pairs are included in the half-reaction, the total change in ox.no. is 2 (1) = 2 units, in agreement with the 2 e in the balanced equation. 4) (CN) OH 2 CNO + 2 H 2 O + 2 e The ox.no. of the atom-pair, (CN), may be considered to change from 0 to +1; oxidation occurs (= loss of electrons); a change of 1 unit per (CN). For the two (CN) atom-pairs in the half-reaction, the total change is therefore 2 (1) = 2 units. This agrees with the 2 e shown in the equation. (Note, in the CNO ion, the ox.no. of the (CN) atom-pair is determined by first assigning the O atom a usual ox.no. of 2 ; then, since the charge on the CNO ion is 1, the (CN) atom-pair must be assigned an ox.no. of +1.) By way of re-emphasis regarding this Assignment: The number of electrons needed to balance a half-reaction is to be decided by using "The Procedure" given at the start of this Assignment. Subsequently, that computation (i.e., the summing of the electrostatic charges and comparing of the sums) may later be checked for accuracy by determining the total change of ox.no. within the balanced half-reaction. (This serves as an "independent" check of accuracy.) Both methods require that ALL ELECTROSTATIC CHARGES MUST BE CORRECTLY WRITTEN. Otherwise, "garbage in, garbage out!"