Download presentation

Presentation is loading. Please wait.

Published byMakayla McConnell Modified over 2 years ago

1
Balancing Redox Equations in Acidic Conditions Take the reaction between potassium permanganate KMnO 4 ) and sodium sulfite (NaSO 3 ) 7 steps are required to balance the full ionic equation from 2 separate half equations (oxidation & reduction)

2
Step 1 Write 2 half-equations for the reaction. MnO 4 - Mn 2+ SO 3 2- SO 4 2-

3
Step 2 Balance oxygen atoms using H 2 O MnO 4 - Mn H 2 O SO H 2 O SO 4 2-

4
Step 3 Balance H with H + 8H + + MnO 4 - Mn H 2 O SO H 2 O SO H +

5
Step 4 Balance the charges with electrons 8H + + MnO e - Mn 2+ 4H 2 O SO H 2 O SO H + + 2e -

6
Step 5 Multiply the 2 half-equations by whole numbers (the lowest common multiple of the 2 stoichiometric coefficients in front of the electrons) so that electrons gained in the reduction reaction equals electrons given out by the oxidation reaction In this case the reduction reaction needs to be multiplied by 2 while the oxidation reaction needs to be multiplied by 5 to get a common number of electrons 10 16H + + 2MnO e - 2Mn 2+ 8H 2 O 5SO H 2 O 5SO H e -

7
Step 6 Add the 2 half-reactions 16H + + 2MnO e - + 5SO H 2 O 2Mn 2+ 8H 2 O + 5SO H e - The electrons cancel on the both sides to give: 16H + + 2MnO SO H 2 O 2Mn 2+ 8H 2 O + 5SO H +

8
Step 7 Subtract H + & H 2 O which occur on both sides of the equation The consumption of 16H + & the production of 10H + is equal to a net consumption of 6H + The consumption of 5 H 2 O molecules & the production of 8 H 2 O molecules is equal to the net production of 3 H 2 O molecules 6H + + 2MnO SO Mn 2+ 3H 2 O + 5SO 4 2-

9
Points to note The H 2 O molecules always appear on the right hand- side (RHS) of the reduction (8H + + MnO e - Mn 2+ 4H 2 O) reaction but on the left-hand side (LHS) of the oxidation (SO H 2 O SO H + + 2e - reaction The H + ions always appear on the opposite side of H 2 O molecules in both of the half-equations and the net ionic equation. 6H + + 2MnO SO Mn 2+ 3H 2 O + 5SO 4 2-

10
Balancing redox equations for neutral or alkaline conditions The reaction taking place is between potassium permanganate and sodium sulfite to form manganese dioxide (MnO 2 ) The method used for balancing equations in acidic conditions is used; then 1 OH - is added for every H + in the equation

11
Step 1 Write the 2 half-reactions MnO 4 - MnO 2 SO 3 2- SO 4 2-

12
Steps 2-7 Follow the same steps as steps 2-4 for the reaction in acidic media to get the following 2 half- equations: 4H + + MnO e - MnO 2 + 2H 2 O (x2) SO H 2 O SO H + + 2e - (x3) Multiplying the reduction reaction by 2, the oxidation by 3, adding together and simplifying gives: 2H + + 2MnO SO MnO 2 + H 2 O + 3SO 4 2-

13
Step 8 Add OH - to convert any H + to H 2 O. Any OH - added to 1 side of the equation must also be added to the other side. 2H + + 2OH - 2MnO SO MnO 2 + H 2 O + 3SO OH - Which simplifies to: 2H 2 O + 2MnO SO MnO 2 + H 2 O + 3SO OH - Which simplifies further to: H 2 O + 2MnO SO MnO 2 + 3SO OH -

14
Balancing Redox equations in strongly alkaline conditions Take the reaction between potassium permanganate & sodium sulfite in strongly alkaline media MnO 4 - MnO 4 2- SO 3 2- SO 4 2- Following the same steps of balancing the equation in acidic media, then adding OH- for every H+ results in: 2MnO SO OH - 2MnO SO H 2 O

15
Redox Titrations Similar to acid-base titrations Acid-base titration: transfer of 1 or more hydrogen ions (protons) from the acid to the base Redox Titration: transfer of one/more electrons from a reducing agent to an oxidizing agent

16
Oxidizing agents for redox titrations Acidified manganate (VII) ions (permanagate) 8H + + MnO e - Mn 2+ 4H 2 O MnO 4 - is purple but Mn 2+ is almost colourless

17
Oxidizing agent for redox titrations 2 Acidified dichromate (VI) ions 14H + + Cr 2 O e - 2Cr H 2 O Cr 2 O 7 2- are orange in colour Cr 3+ is green Can be used as primary standards (a reagent which is very pure, & can be used to prepare a solution of known concentration)

18
Some more oxidizing agents Iron (III) ions/salts Fe 3+ + e - Fe 2+ Iodine: I 2 + 2e - 2I - I 2 is red brown while I - is colourless Acidified hydrogen peroxide H 2 O 2 + 2H + + 2e - 2H 2 O

19
Reducing agents for redox titrations Iron(II) salts/ions Fe 2+ Fe 3+ + e - Hydrogen peroxide if a more powerful oxidizing agent e.g. dichromate (VI) or manganate (VII) is present H H + + O 2 + 2e - Iodide ions: 2I - I 2 + 2e - Sodium thiosulfate (VI): 2S 2 O 3 2- S 4 O e-

Similar presentations

© 2016 SlidePlayer.com Inc.

All rights reserved.

Ads by Google