Answers

1. Ni + 2(-2) = 0

Ni + -4     = 0

           Ni =4

2. U + 2(-2) =2

U + -4    = 2

         U  = 6

3. 3(1) + N = 0         N + 2(-2)=0

3      + N =0          N + -4    =0

           N = -3                 N  =4

4. Mn²⁺ + NaBiO₃ ® Bi³⁺ + MnO₄^-

Mn + 4(-2) = -1          1 + Bi + 3(-2) =0

   Mn + -8  =-1             1 + Bi + -6   =0

           Mn = 7                -5    + Bi   =0

                                                   Bi=5

                            +7

             Mn²⁺ ® MnO₄^- +5e-

                           +5

             2e- + NaBiO₃® Bi³⁺

             Mn²⁺ is the reducing agent because it donated electrons to MnO₄^- to reduce the

             Oxidation state of one of its atoms(Mn)

             Bi³⁺ is the oxidizing agent because it has accepted electrons from another

              Reactant

5. CH₄ + 2O₂ ® CO₂ + 2H₂O

C + 4(1) =0       C + 2(-2)=0            O=0       Oxygens on right side= -8

 C + 4    = 0        C + -4 = 0

        C  = -4              C  =4

             -4         +4

             CH₄ ® CO₂ + 8e-

                          0                -8

              8e- + 2O₂ ® CO₂ + 2H₂O

              Carbon is oxidized (there has been a decrease in its oxidation state

6. Zn + 2HCl ® ZnCl₂ + H₂

Zn = 0    Zn + 2(-1) =0         H + -1 =0       H =0

                Zn   + -2  = 0                H  =1

                            Zn =2

0        +2

Zn ® ZnCl₂

+1                      0

HCl ® ZnCl₂ + H₂

HCl(H) has been reduced because there has been an increase in oxidation state.

7. 2CuCl ® CuCl₂ + Cu

Cu + -1 =0   Cu + 2(-1) =0      Cu=0

        Cu=1      Cu + -2  =0

                               Cu  = 2

   +1         +2

2CuCl ® CuCl₂ +1e-

   +1                        0

        1e- +2CuCl ® CuCl₂ + Cu

CuCl is the oxidizing agent and the reducing agent because it both gives and takes electrons to change the oxidation states of the two Cu on the right side

     8.2H₂O₂ ® 2H₂O + O₂

         2(1) + 2 O = 0        2(1) + O = 0     O = 0

           2 +    2 O = 0           2 + O  = 0

                    2 O = -2              O    = -2

                       O = -1

                -1                     -2

          2H₂O₂ +1e- ® 2H₂O

                -1         0

          2H₂O₂  ® O₂ + 1e-

           H₂O₂ ( O) has been oxidized and reduced because it has both gained and lost   electrons

     9. Al ® AlO₂^-

         NO₂^- ® NH₃

    10. Ag ® Ag(CN)₂^-

          CN^- ® Ag(CN)₂^-

    11. Br ^–_(aq) + MnO₄^- _(aq)® Br_2(l) + Mn²⁺_(aq)

            Br=-1  Br =2    Mn + 4(-2) = -1      Mn = 2

                                            Mn + -8 = -1

                                                  Mn  = 7                      

         2Br ^- ® Br₂ + 2e-

         MnO₄ ^- ® Mn²⁺

         5(2Br- ® Br₂ + 2e-)

         2(8H⁺ + MnO₄^- + 5e- ® Mn + 4H₂O)

         10Br^- ® 5Br₂ + 10e-

      +  16H⁺ + 2MnO₄^- + 10e- ® 2Mn + 8H₂O

      16H⁺_(aq) + 2MnO₄^-_(aq) + 10Br^-_(aq) ® 5Br_2(l) + 2Mn²⁺_(aq) + 8H₂O_(l)

     12. Cr_(s) + CrO₄^2-_(aq) ® Cr(OH)_3(s)

           Cr = 0    Cr + 4(-2) = -2      Cr + 3(-2) + 3(1) =0

                           Cr + -8   = -2        Cr + -6 + 3        = 0

                                    Cr = 6               Cr + -3       = 0

                                                                          Cr = 3

           Cr ® Cr(OH)₃ +3e-

      3e- + CrO₄^- ® Cr(OH)₃

          2H₂O

         3H₂O + Cr ® Cr(OH)₃ + 3e- + 3H⁺

         2H⁺

       + 5H⁺ + 3e- + CrO₄^2- ® Cr(OH)₃ + H₂O

     2H⁺ +2OH^-  + 2H₂O + Cr + CrO₄^2- ® 2Cr(OH)₃ + 2OH^-

        2H₂O

    4H₂O_(l) + Cr_(s) + CrO₄^2-_(aq) ® 2Cr(OH)_3(s) + 2OH^-_(aq)

     13. Cr₂O₇^2-_(aq) + Cl^-_(aq) ® Cr³⁺_(aq) + Cl_2(g)

           2Cr + 7(-2) = -2        Cr=3     Cl=-1    Cl=0

             2Cr + -14 = -2

                      2Cr  =12

                        Cr = 6

                6e- + Cr₂O₇^2- ® Cr³⁺

                 Cl^- ® Cl₂ + 2e-

               14H⁺ + Cr₂O₇^2- + 6e-® 2Cr³⁺ + 7H₂O

               3 ( 2Cl^-  ® Cl₂ + 2e-)

                14H⁺ + Cr₂O₇^2- + 6e-® 2Cr³⁺ + 7H₂O

              +  6Cl- ® 3Cl₂ +6e-

                14H⁺_(aq) + Cr₂O₇^2-_(aq) + 6Cl^-_(aq) ® 3Cl_2(g) + 2Cr³⁺_(aq) + 7H₂O_(l)