Balance the following equation in acidic condition...

Question

Balance the following equation in acidic conditions. Phases are optional. Cr2 + H2MoO4 ::> Cr3  + Mo

Jules Bruno · Accepted Answer

Balance the redox reaction under acidic conditions:Cr2+(aq)  + H2MoO4(aq) → Cr3+(aq) + Mo(s) *Before balancing the redox reaction:• ignore H+, OH-, and H2O in the starting reactionStep 1: Separate the whole reaction into two half-reactionsCr2+(aq)  →  Cr3+(aq)                                                    H2MoO4(aq) → Mo(s) Step 2: Balance the non-hydrogen and non-oxygen elements firstCr2+(aq)  →  Cr3+(aq)                                                    H2MoO4(aq) → Mo(s) (Cr: balanced)                                               (Mo: balanced)Step 3: Balance oxygen by adding H2O(l) to the side that needs oxygen (1 O : 1 H2O)Cr2+(aq)  →  Cr3+(aq)                                                    H2MoO4(aq) → Mo(s) + 4H2O(l)       (No oxygens)                                               [readmore]Step 4: Balance hydrogen by adding H+ to the side that needs hydrogen (1 H: 1 H+)Cr2+(aq)  →  Cr3+(aq)                                                    6H+ (aq) + H2MoO4(aq) → Mo(s) + 4H2O(l)      (No hydrogens)Step 5: Balance the charges: add electrons to the more positive side (or less negative side) ▪ For species that are neutral: oxidation state = 0▪ For ions (or species with charges): oxidation state = charge of the ion▪ The reactant side and the product side should have an equal overall charge (after adding electrons)Cr2+(aq)  →  Cr3+(aq)                                          6H+ (aq) + H2MoO4(aq) → Mo(s) + 4H2O(l)  Reactants                            Products                               Reactants                            Products                               Cr2+ = +2                                   Cr3+ = +3                               H2MoO4 = 0                       Mo = 0                                                     +1 e- = -1                                 6H+ = 6(+1) = + 6                H2O = 0-----------------------                 ------------------------                     +6 e- = (-1) x6 = -6overall = +2                             overall = +2                             -------------------------------      ----------------                      overall = 0                           overall = 0Step 6: Balance electrons on the two half-reactions▪ multiply the reactions so both half-reactions have the same number of electrons▪ electrons should cancel in the overall reaction            Cr2+(aq)  →  Cr3+(aq) + 1 e- → multiply by 6: 6x [Cr2+(aq)  →  Cr3+(aq) + 1 e- ] 6H+ (aq) + H2MoO4(aq) + 6 e- → Mo(s) + 4H2O(l)  ▪ Oxidation half-reaction: lost electrons (electrons are on the product side) ▪ Reduction half-reaction: gained electrons (electrons are on the reactant side)Step 7: Get the overall reaction by adding the two reaction.▪ species that appear in both reactions on opposite sides are subtracted                    6Cr2+(aq)  →  6Cr3+(aq) + 6 e-  6H+ (aq) + H2MoO4(aq) +6 e- →  Mo(s) + 4H2O(l)  --------------------------------------------------------  6Cr2+(aq) + 6H+ (aq) + H2MoO4(aq)→6Cr3+(aq)  +  Mo(s) + 4H2O(l)  Balanced Redox Reaction:           6Cr2+(aq) + 6H+ (aq) + H2MoO4(aq)→ 6Cr3+(aq)  +  Mo(s) + 4H2O(l)

Problem: Balance the following equation in acidic conditions. Phases are optional. Cr2 + H2MoO4 ::> Cr3 + Mo

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Problem: Balance the following equation in acidic conditions. Phases are optional. Cr2 + H2MoO4 ::&gt; Cr3 + Mo

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Problem: Balance the following equation in acidic conditions. Phases are optional. Cr2 + H2MoO4 ::> Cr3 + Mo