Redox Reactions -- Electrons Transferred?

Question

Redox Reactions -- Electrons Transferred

Balance each of the following equations occuring in basic aqueous solution.

In each case enter the number (n) of electrons transferred from the reducing agent to the oxidizing agent for the conventionally balanced equation (full reaction).

Al + NO3- NH3 + Al(OH)4- n = 24
CrO42- + PH3 Cr(OH)4- + P n = 3
CO32- + N2H4 CO + N2 n = ?
SO32- + ZrO(OH)2 SO42- + Zr n = ?

No idea on the last too... anyone willing to help me?

Answer 1

1)
ox: Al + 4 OH- ---> Al(OH)4- + 3 e-
red: NO3- + 8 e- + 6 H2O ----> NH3 + 9 OH-
overall: 8 Al + 5 OH- + 3 NO3- + 18 H2O ---> 8 Al(OH)4- + 3 NH3 n=24

2)
ox: PH3 + 3 OH- ---> P + 3 e- + 3H2O
red: CrO4(2-) + 3 e- + 4 H2O---> Cr(OH)4- + 4 OH-
overall: PH3 + CrO4(2-) + H2O ---> Cr(OH)4- + P + OH- n=3

3)
ox: N2H4 + 4 OH- ---> N2 + 4 e- + 4 H2O
red: CO3(2-) + 2 e- + 2 H2O ----> CO + 4 OH-
overall: N2H4 + 2 CO3(2-) ---> N2 + 2 CO + 4 OH- n=4

4)
ox: SO3(2-) + 2 OH- ---> SO4(2-) + 2 e- + H2O
red: ZrO(OH)2 + 4 e- + H2O ----> Zr + 4 OH-
overall: 2 SO3(2-) + ZrO(OH)2 ---> 2 SO4(2-) + Zr + H2O n=4

Answer 2

well, i shall try my best....


For the first one:
ox: Al + 4 OH- ---> Al(OH)4- + 3 e-
red: NO3- + 8 e- + 6 H2O ----> NH3 + 9 OH-
overall: 8 Al + 5 OH- + 3 NO3- + 18 H2O ---> 8 Al(OH)4- + 3 NH3 n=24

For the second one:
ox: PH3 + 3 OH- ---> P + 3 e- + 3H2O
red: CrO4(2-) + 3 e- + 4 H2O---> Cr(OH)4- + 4 OH-
overall: PH3 + CrO4(2-) + H2O ---> Cr(OH)4- + P + OH- n=3

For the third one :
ox: N2H4 + 4 OH- ---> N2 + 4 e- + 4 H2O
red: CO3(2-) + 2 e- + 2 H2O ----> CO + 4 OH-
overall: N2H4 + 2 CO3(2-) ---> N2 + 2 CO + 4 OH- n=4

For the last question :
ox: SO3(2-) + 2 OH- ---> SO4(2-) + 2 e- + H2O
red: ZrO(OH)2 + 4 e- + H2O ----> Zr + 4 OH-
overall: 2 SO3(2-) + ZrO(OH)2 ---> 2 SO4(2-) + Zr + H2O n=4

Hope this helps you..........

Answer 3

MnO4- + Zn ⇒ Zn(OH)2 + MnO2 4H2O +2MnO4- + 3Zn⇒ 2MnO2 + 3Zn(OH)2 +2OH- SnO2^2- + Bi(OH)3➝Bi + SnO3^2- 3SnO2^2- + 2Bi(OH03➝3SnO3^2- +2Bi + 3H2O OCL^- + Cr2O3➝ CrO4^2- + Cl^- 4OH^- +3OCL^- + Cr2O3➝3Cl^- +2CrO4^2- +2H2O