Balance the equation for redox reactions. Show as a net ionic equation. Identify

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N2H4 doesn't get split up because it's not an ionic compound. Your basic half-reactions are OK N2H4 ==> N2 . . . . . . .and . . . . . .BrO3- ==> Br- Look at BrO3-. Since the oxidation state of O is -2, then Br must be +5 (5 - 6 = -1, the charge on BrO3-). On the right side, Br is -1. To go from +5 to -1 means gaining 6 electrons; that is reduction). Likewise, in N2H4, H is +1, so 4H = +4; Since N2H4 has no charge overall, 2N must equal -4, so N = -2. On the right, the oxidation state for N is 0. To go from -2 to 0, N lost 2 electrons; that is oxidation. Balance each half-reaction. I disagree with your instructor about not using H+. BrO3- ==> Br- . . .add 3 H2O to the right side to balance oxygen BrO3- ==> Br- + 3H2O . . .now add 6H+ to the left to balance H BrO3- + 6H+ ==> Br- + 3H2O . . .add 6e- to the left to balance the charge BrO3- + 6H+ + 6e- ==> Br- + 3H2O (Equation 1) N2H4 ==> N2 . . .add 4H+ to the right to balance H N2H4 ==> N2 + 4H+ . . .add 4e- to the right to balance the charge N2H4 ==> N2 + 4H+ + 4e- (Equation 2) The first half reaction (with the Br) has 6e- on the left and the second half-reaction (with N) has 4e- on the right. The least common multiple of 6 and 4 is 12. So, to cancel the electrons, multiply Equation 1 by 2 and multiply equation 2 by 3; add the two equations together. 2BrO3- + 12H+ + 12e- ==> 2Br- + 6H2O 3N2H4 ==> 3N2 + 12H+ + 12e- =================================== 2BrO3- + 3N2H4 ==> 2Br- + 3N2 + 6H2O

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