Balancing redox reactions in acidic solution - step 1Cr₂O₇²⁻ + Fe²⁺ExampleCr³⁺ + Fe³⁺From this question, we mustdivide them into half reactionby determine which one isoxidation or reduction. You cancalculate the oxidation numberfirst to make this step easy.For Fe, the oxidationnumber changes from+2 to +3 → oxidation.For Cr, we mustcalculate the oxidationnumber first.Cr₂O₇²⁻2Cr + 7(-2) = -22Cr - 14 = -22Cr = 12Cr = +6So Cr changes from +6to +3 → reductionCr₂O₇²⁻ + Fe²⁺ Cr³⁺ + Fe³⁺+6 +2 +3 +3Oxidation Fe²⁺ Fe³⁺Reduction Cr₂O₇²⁻ Cr³⁺After this step, nowwe can balance eachhalf reaction.
Balancing redox reactions in acidic solution - step 2Oxidation Fe²⁺ Fe³⁺ + e⁻ReductionCr₂O₇²⁻ Cr³⁺When balancing thehalf-equation, makesure the total chargeon both sides is equal.Left = +2Right = +3 + (–1) = +2BalancedFor this half-equation, balance the element other than O and H. There are 2 Cratoms in Cr₂O₇²⁻, so we need to add 2 at Cr³⁺ on the right-hand side.Cr₂O₇²⁻2Cr³⁺Cr₂O₇²⁻2Cr³⁺ + 7H₂O In acidic solution, balance oxygen by adding H₂O molecules to the right side. Inthis example, Cr₂O₇²⁻ has 7 oxygen atoms. So balance the oxygen by adding 7 H₂Omolecules to the right side.Cr₂O₇²⁻ + 14H⁺ 2Cr³⁺ + 7H₂O Balance the hydrogen atom. In this example, there are 14 hydrogen atoms in7H₂O. So, add 14 H⁺ to the left side.Cr₂O₇²⁻ + 14H⁺ + 6e⁻ 2Cr³⁺ + 7H₂O Lastly, balance the charge. So we need to check the total charge on both sides.Left side: -2 + 14(+1) = +12Right side: 2(+3) + 7(0) = +6To make the charges equal, add 6 electrons (6e⁻) to the left side. Since this is areduction, electrons are gained.Try to understand andremember all this step ifyou find a question likethis.
Balancing redox reactions in acidic solution - step 2 & 36Fe³⁺ + 6e⁻Cr₂O₇²⁻ + 14H⁺ + 6e⁻ 2Cr³⁺ + 7H₂O6Fe²⁺ + Cr₂O₇²⁻ + 14H⁺// 6Fe³⁺ + 2Cr³⁺ + 7H₂O(Fe²⁺ Fe³⁺ + e⁻)After balance all the halfequation, multiply each halfequation by an integer ifnecessary in order tocombine the half-equations. x 6Cr₂O₇²⁻ + 14H⁺ + 6e⁻ 2Cr³⁺ + 7H₂O To balance the number of electrons:Multiply the oxidation equation by 6, because the reduction involves 6 electrons.Multiply the reduction equation by 1 (since it already has 6 electrons total).6Fe²⁺Combine the half reactionElectrons are cancelled outIn the oxidation half, 6 electrons are released.In the reduction half, 6 electrons are gained.Since both sides involve the same number ofelectrons (6e⁻), they cancel each other whenthe half-equations are added.Always remember tocombine the remainingspecies by combine all thereactants on the left sideand all the products on theright side.Make sure to do a finalcheck on atoms andcharges