What are the various methods to balance a redox reaction?

Oxidation-Reduction Reactions, or redox reactions, are reactions in which one reactant is oxidized and one reactant is reduced simultaneously. This module demonstrates how to balance various redox equations. Balancing Redox Reactions Balancing redox reactions is slightly more complex than balancing standard reactions, but still follows a relatively simple set of rules. One major difference is the necessity to know the half-reactions of the involved reactants; a half-reaction table is very useful for this. Half-reactions are often useful in that two half reactions can be added to get a total net equation. Although the half-reactions must be known to complete a redox reaction, it is often possible to figure them out without having to use a half-reaction table. This is demonstrated in the acidic and basic solution examples. Besides the general rules for neutral conditions, additional rules must be applied for aqueous reactions in acidic or basic conditions. The method used to balance redox reactions is called the Half Equation Method. In this method, the equation is separated into two half-equations; one for oxidation and one for reduction. Each equation is balanced by adjusting coefficients and adding H2O, H+, and e- in this order: Balance elements in the equation other than O and H. Balance the oxygen atoms by adding the appropriate number of water (H2O) molecules to the opposite side of the equation. Balance the hydrogen atoms (including those added in step 2 to balance the oxygen atom) by adding H+ ions to the opposite side of the equation. Add up the charges on each side. Make them equal by adding enough electrons (e-) to the more positive side. (Rule of thumb: e- and H+ are almost always on the same side.) The e- on each side must be made equal; if they are not equal, they must be multiplied by appropriate integers (the lowest common multiple) to be made the same. The half-equations are added together, canceling out the electrons to form one balanced equation. Common terms should also be canceled out. (If the equation is being balanced in a basic solution, through the addition of one more step, the appropriate number of OH- must be added to turn the remaining H+ into water molecules.) The equation can now be checked to make sure that it is balanced. Balancing in a Neutral Solution:- Balance the following reaction Cu+(aq)+Fe(s)?Fe3+(aq)+Cu(s) Step 1: Separate the half-reactions. By searching for the reduction potential, one can find two separate reactions: Cu+(aq)+e??Cu(s) Fe3+(aq)+3e??Fe(s) The copper reaction has a higher potential and thus is being reduced. Iron is being oxidized so the half-reaction should be flipped. This yields: Cu+(aq)+e??Cu(s) Fe(s)?Fe3+(aq)+3e? Step 2: Balance the electrons in the equations. In this case, the electrons are simply balanced by multiplying the entire Cu+(aq)+e??Cu(s) half-reaction by 3 and leaving the other half reaction as it is. This gives: 3Cu+(aq)+3e??3Cu(s) Fe(s)?Fe3+(aq)+3e? Step 3: Adding the equations give: 3Cu+(aq)+3e?+Fe(s)?3Cu(s)+Fe3+(aq)+3e? The electrons cancel out and the balanced equation is left. 3Cu+(aq)+Fe(s)?3Cu(s)+Fe3+(aq) Acidic Conditions:- Balance the following redox reaction in acidic conditions. Cr2O2?7(aq)+HNO2(aq)?Cr3+(aq)+NO?3(aq) Step 1: Separate the half-reactions. The table provided does not have acidic or basic half-reactions, so just write out what is known. Cr2O2?7(aq)?Cr3+(aq) HNO2(aq)?NO?3(aq) Step 2: Balance elements other than O and H. In this example, only chromium needs to be balanced. This gives: Cr2O2?7(aq)?2Cr3+(aq) HNO2(aq)?NO?3(aq) Step 3: Add H2O to balance oxygen. The chromium reaction needs to be balanced by adding 7 H2O molecules. The other reaction also needs to be balanced by adding one water molecule. This yields: Cr2O2?7(aq)?2Cr3+(aq)+7H2O(l) HNO2(aq)+H2O(l)?NO?3(aq) Step 4: Balance hydrogen by adding protons (H+). 14 protons need to be added to the left side of the chromium reaction to balance the 14 (2 per water molecule * 7 water molecules) hydrogens. 3 protons need to be added to the right side of the other reaction. 14H+(aq)+Cr2O2?7(aq)?2Cr3+(aq)+7H2O(l) HNO2(aq)+H2O(l)?3H+(aq)+NO?3(aq) Step 5: Balance the charge of each equation with electrons. The chromium reaction has (14+) + (2-) = 12+ on the left side and (2 * 3+) = 6+ on the right side. To balance, add 6 electrons (each with a charge of -1) to the left side: 6e?+14H+(aq)+Cr2O2?7(aq)?2Cr3+(aq)+7H2O(l) For the other reaction, there is no charge on the left and a (3+) + (-1) = 2+ charge on the right. So add 2 electrons to the right side: HNO2(aq)+H2O(l)?3H+(aq)+NO?3(aq)+2e? Step 6: Scale the reactions so that the electrons are equal. The chromium reaction has 6e- and the other reaction has 2e-, so it should be multiplied by 3. This gives: 3?[HNO2(aq)+H2O(l)?3H+(aq)+NO?3(aq)+2e?]? 3HNO2(aq)+3H2O(l)?9H+(aq)+3NO?3(aq)+6e? 6e?+14H+(aq)+Cr2O2?7(aq)?2Cr3+(aq)+7H2O(l). Step 7: Add the reactions and cancel out common terms. [3HNO2(aq)+3H2O(l)?9H+(aq)+3NO?3(aq)+6e?]+ [6e?+14H+(aq)+Cr2O2?7(aq)?2Cr3+(aq)+7H2O(l)]= 3HNO2(aq)+3H2O(l)+6e?+14H+(aq)+Cr2O2?7(aq)?9H+(aq)+3NO?3(aq)+6e?+2Cr3+(aq)+7H2O(l) The electrons cancel out as well as 3 water molecules and 9 protons. This leaves the balanced net reaction of: 3HNO2(aq)+5H+(aq)+Cr2O2?7(aq)?3NO?3(aq)+2Cr3+(aq)+4H2O(l) Basic Conditions:- Balance the following redox reaction in basic conditions. Ag(s)+Zn2+(aq)?Ag2O(aq)+Zn(s) Go through all the same steps as if it was in acidic conditions. Step 1: Separate the half-reactions. Ag(s)?Ag2O(aq) Zn2+(aq)?Zn(s) Step 2: Balance elements other than O and H. 2Ag(s)?Ag2O(aq) Zn2+(aq)?Zn(s) Step 3: Add H2O to balance oxygen. H2O(l)+2Ag(s)?Ag2O(aq) Zn2+(aq)?Zn(s) Step 4: Balance hydrogen with protons. H2O(l)+2Ag(s)?Ag2O(aq)+2H+(aq) Zn2+(aq)?Zn(s) Step 5: Balance the charge with e-. H2O(l)+2Ag(s)?Ag2O(aq)+2H+(aq)+2e? Zn2+(aq)+2e??Zn(s) Step 6: Scale the reactions so that they have an equal amount of electrons. In this case, it is already done. Step 7: Add the reactions and cancel the electrons. H2O(l)+2Ag(s)+Zn2+(aq)?Zn(s)+Ag2O(aq)+2H+(aq). Step 8: Add OH- to balance H+. There are 2 net protons in this equation, so add 2 OH- ions to each side. H2O(l)+2Ag(s)+Zn2+(aq)+2OH?(aq)?Zn(s)+Ag2O(aq)+2H+(aq)+2OH?(aq). Step 9: Combine OH- ions and H+ ions that are present on the same side to form water. H2O(l)+2Ag(s)+Zn2+(aq)+2OH?(aq)?Zn(s)+Ag2O(aq)+2H2O(l) Step 10: Cancel common terms. 2Ag(s)+Zn2+(aq)+2OH?(aq)?Zn(s)+Ag2O(aq)+H2O(l) read less

Break the original equation into two equation- Oxidation & Reduction; Balance elements in the equation other than O and H; Balance the oxygen atoms by adding the appropriate number of water (H2O) molecules to the opposite side of the equation; Balance the hydrogen atoms (including those added in step 2 to balance the oxygen atom) by adding H+ ions to the opposite side of the equation; Add up the charges on each side. Make them equal by adding enough electrons (e-) to the more positive side. (Rule of thumb: e- and H+ are almost always on the same side.); The e- on each side must be made equal; if they are not equal, they must be multiplied by appropriate integers (the lowest common multiple) to be made the same; The half-equations are added together, cancelling out the electrons to form one balanced equation. Common terms should also be cancelled out. read less