BALANCING REDOX REACTIONS

Transcription

1 BALANCING REDOX REACTIONS Review of Titration: o What is a titration? Transfer of a solution from a burette into a measured volume of a sample solution. o What do you need to know in order to carry out a titration? Volume of aliquot Concentration of titrant solution Indicator of endpoint BALANCED EQUATION of the reaction that is occurring to know the molar relationship o How can titrations be applied to redox chemistry? Like video said: we can oxidize or reduce species For example, we want to oxidize Fe 2+ (aq) Fe 3+ (aq) + e Using Potassium Permanganate o Where can we look to find the halfreaction that potassium permanganate undergoes? TABLE OF STANDARD REACTION POTENTIALS!! MnO 4 (aq) + 8 H + (aq) + 5 e Mn 2+ (aq) + 4 H 2 O (l)

2 o Now, how do we balance these two half reactions to form one net ionic equation? THREE METHODS OF BALANCING REDOX REACTIONS: 1. Inspection 2. Table of Standard Reduction Potentials 3. HalfReaction 1. INSPECTION (Balance for charge) o When the redox reaction is simple and you can clearly see how to balance out the charges so both sides carry the same charge you can do so. Al + Cu 2+ Al 3+ + Cu Charge on Left = + 2 Charge on Right = +3 In order to balance charges, you must have: Al + 3 Cu 2+ 2 Al 3+ + Cu Charge on Left = + 6 Charge on Right = + 6 Now, to balance the atoms, the equation becomes: 2 Al + 3 Cu 2+ 2 Al Cu 2. TABLE OF STANDARD REDUCTION POTENTIAL o In the event that the reaction is too complex to solve by simple inspection, you an use the halfreactions in the SRP table and

3 balance the halfreactions so they will be electrically neutral when added together (balance electrons). Previous Titration Example: MnO 4 (aq) + 8 H + (aq) + 5 e Mn 2+ (aq) + 4 H 2 O (l) Fe 2+ (aq) Fe 3+ (aq) + e To balance out the electrons, we can multiply the Iron halfreaction by 5 MnO 4 (aq) + 8 H + (aq) + 5 e Mn 2+ (aq) + 4 H 2 O (l) 5 Fe 2+ (aq) 5 Fe 3+ (aq) + 5 e MnO 4 (aq) + 8 H + (aq) + 5 Fe 2+ (aq) Mn 2+ (aq) + 4 H 2 O (l) + 5 Fe 3+ (aq) 3. HALFREACTION o When halfreactions are not found within the SRP table, we must balance them by the halfreaction method o This method is different in ACIDIC and BASIC conditions ACIDIC presence of H + BASIC presence of OH Steps for Balancing a Redox Reaction under ACIDIC Conditions:

4 STEP 1: Break reaction into unbalanced halfreactions according to similar species present. Each halfreaction must be balanced separately. P 4 + IO 3 H 2 PO 4 + I BROKEN INTO: P 4 H 2 PO 4 IO 3 I STEP 2: Balance Atoms IO 3 I STEP 3: Balance for Oxygen Using H 2 O IO 3 I + 3 H 2 O STEP 4: Balance for Hydrogen using H + 6 H + + IO 3 I + 3 H 2 O STEP 5: Balance for Charge using Electrons

5 6 H + + IO 3 I + 3 H 2 O Charge on Left = + 5 Charge on Right = 1 So, to balance out, we should add SIX ELECTRONS to the left side. 6 e + 6 H + + IO 3 I + 3 H 2 O Charge on Left = 1 Charge on Right = 1 STEP 6: Balance Other HalfReaction and Add Them Together P 4 H 2 PO 4 2. P 4 4 H 2 PO H 2 O + P 4 4 H 2 PO H 2 O + P 4 4 H 2 PO H H 2 O + P 4 4 H 2 PO H e NOW TO ADD: 6 e + 6 H + + IO 3 I + 3 H 2 O 16 H 2 O + P 4 4 H 2 PO H e

6 At this point, we must make sure that the ELECTRONS will cancel out when we ADD the reactions together, so we have to multiply the top by 10 and the bottomr by e + 60 H IO 3 10 I + 30 H 2 O 48 H 2 O + 3 P 4 12 H 2 PO H e 60 H IO H 2 O + 3 P 4 10 I + 30 H 2 O + 12 H 2 PO H + Now, we can cancel out hydrogen and water that are on both sides to simplify. 10 IO H 2 O + 3 P 4 10 I + 12 H 2 PO H + ***** CHECK:****** A great way to make sure you have balanced correctly is to check the charge balance on both sides of the reaction. Charge on Left Side = 10 Charge on Right Side = (10) + (12) + (12) = 10 Balancing a Reaction in BASIC Conditions: Repeat STEPS 14 for Redox Reactions Occurring in Acidic Solution STEP 5: Count the number of H + ions in the equation and add the same number of OH ions TO BOTH SIDES. On a side that has both H + and OH ions, they will combine to form H 2 O which can then be cancelled out accordingly. STEP 6: Balance for Charge Using Electrons

7 STEP 7: Balance the Other Half Reaction and Add the Two together ******CHECK:******** Remember to check for Charge Balance EXAMPLE: MnO 4 + CN MnO 2 + CNO (in base) STEP 1: MnO 4 MnO 2 CN CNO STEP 2: MnO 4 MnO 2 STEP 3: MnO 4 MnO H 2 O STEP 4: 4 H + + MnO 4 MnO H 2 O STEP 5: 4 OH + 4 H + + MnO 4 MnO H 2 O + 4 OH 4 H 2 O (4 OH + 4 H + ) + MnO 4 MnO H 2 O + 4 OH 4 H 2 O + MnO 4 MnO H 2 O + 4 OH 2 H 2 O + MnO 4 MnO OH STEP 6: 3 e + 2 H 2 O + MnO 4 > MnO OH

8 STEP 7: H 2 O + CN CNO H 2 O + CN CNO + 2 H + 2 OH + H 2 O + CN CNO + 2 H OH 2 OH + H 2 O + CN CNO + 2 H OH (2 H 2 O) 2 OH + H 2 O + CN 2 H 2 O + CNO 2 OH + CN CNO + H 2 O + 2 e Now to ADD: 2 H 2 O + MnO e MnO OH 2 OH + CN CNO + H 2 O + 2 e Multiply the top by two and the bottom by three: 4 H 2 O + 2 MnO e 2 MnO OH 6 OH + 3 CN 3 CNO + 3 H 2 O + 6 e H 2 O + 2 MnO CN 2 MnO OH + 3 CNO