Fe2+ + Cr2o72- = Fe3+ + Cr3+ how do I balance this reaction in (a) acidic soluti

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reduction half equation Cr2O7^2- + 14H+ + 6e ---------> 2Cr^3+ + 7H2O oxidation half equation Fe2+ ----------> Fe3+ + e There are 6 e transferred in the reduction half equation and only 1 in the oxidation half equation. So you will need 6x the oxidation half equation to balance the reduction half equation 6Fe2+ --------> 6Fe3+ + 6e So add the two half equations together 6Fe2+(aq) + Cr2O7^2- + 14H+ ---------> 6Fe3+(aq) + 2Cr3+ + 7H2O The answer is E. - none of these. The coefficient is 6 ps.... here is how to balance the Cr2O7^2- in acidic solution if you want to know. You know that Cr2O7^2- is going to Cr3+ Cr2O7^2- -------> 2Cr3+ balance the O in the Cr2O7^2- by adding H2O to the other side Cr2O7^2- ----------> 2Cr3+ + 7H2O Now balance the H you added in the H2O by adding H+ to the LHS (you can do this because it is in acidic solution) Cr2O7^2- + 14H+ -----> 2Cr3+ + 7H2O Now the atoms all balance but we need to work out the electron transfer SUM up the total charge on each side of the arrow LHS = 1 x (Cr2O7^2-) + 14H+ = 12+ total charge RHS = 2 x Cr3+ = 6+ total charge Now add as many electrons to the MOST positive side as are needed to make the total charge on both sides even. In this case add 6e to the LHS LHS = 12+ + 6e = 6+ = RHS Cr2O7^2- + 14H+ + 6e ---------> 2Cr3+ + 7H2O done, (electrons are being gained so this is the reduction half equation)

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