Balance the reaction between N_2 and Br^- to form Br_2 and N_2H_4 in basic solut

Question

Balance the reaction between N_2 and Br^- to form Br_2 and N_2H_4 in basic solution. When you have balanced the equation using the smallest integers possible, enter the coefficients of the species shown. Enter "1" if the coefficient is "1". _______N_2 + __________Br^- rightarrow________ Br_2+___________N_2H_4 Water appears in the balanced equation as a _____________(reactant, product, neither) with a coefficient of _______.(Enter 0 for neither.) How many electrons are transferred in this reaction? _________

Explanation / Answer

split equations

N2 --> N2H4

Br- = Br2

balance Br

N2 --> N2H4

2Br- = Br2

balance H

4H+ + N2 --> N2H4

2Br- = Br2

balance charges

4e- + 4H+ + N2 --> N2H4

2Br- = Br2 + 2e-

balance e-

4e- + 4H+ + N2 --> N2H4

4Br- = 2Br2 + 4e-

add all

4Br- + 4e- + 4H+ + N2 --> N2H4 + 2Br2 + 4e-

cancel e-

4Br- + 4H+ + N2 --> N2H4 + 2Br2

conver H+ acidic to basic OH-

4OH- + 4Br- + 4H+ + N2 --> N2H4 + 2Br2 + 4OH-

4Br- + 4H2O + N2 --> N2H4 + 2Br2 + 4OH-

then...

water appears in the equation as a REACTANT ( left side )

with a coefficient of 4

4 moles of electrons are being transferred in the reaction

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