3 Redox Half Reactions?

Question

(1.) H+ + NO3¯ + H2S → H2O + NO + S

(2.) Cr2O7¯² + I¯ + H+ → Cr+³ + I2 + H2O

(3.) IO4¯ + I¯ + H+ → I2 + H2O

Answer 1

2 H+ + 2 NO3¯ + 3 H2S → 4 H2O + 2 NO + 3 S

2I- -> I2 + 2e-

Cr2O7(2-) +6e- + 14 H+ -> 2Cr3+ 7H2O

Cr2O7¯² + 6 I¯ + 14 H+ → 2 Cr+³ + 3 I2 + 7 H2O
(Balanced)

2IO4- + 14e- + 16H+ -> I2 + 8H2O

2 IO4¯ + 28 I¯ + 16 H+ → 15 I2 + 8 H2O (Balanced)

Answer 2

2 NO3(-) + 3 H2S + 2 H+ ==> 2 NO + 3 S + 4 H2O

Cr2O7(2-) + 6 I- + 14 H+ ==> 2 Cr3+ + 3 I2 + 7 H2O

2 IO4(-) + 14 I- + 16 H+ ==> 8 I2 + 8 H2O

Answer 3

(1) 4H+ + NO3(-) + 3e = NO + 2H2O
H2S = S + 2e + 2H+
(2) Cr2O7(2-) + 6e +14H+ = 2Cr3+ + 7H2O
2I- = I2 + 2e
(3) 2IO4(-) +16H+ +14e = I2 + 8H2O
2I- =I2+ 2e

These are half reactions, when you balance the number of electrons transfered, you'll get the final equation with the right number of moles

Answer 4

Assign oxidation numbers to each species interior the reaction. a million. Ox..+0, , , +0, , , , ,+a million.-2 . . .2H2 + O2 --> 2H2O H2 is going from +0 to +a million. O2 is going from +0 to -2. So the a million/2-reactions are: 2(H2 ---> H+ + e-) = 2H2 ---> 2H+ + 2e- (a million/2)O2 + 2e- ---> O-- = O + 2e- ---> O-- 2. Ox. .+0. . . .+2. . . . . . .+2.. . . . .0 . . . .Fe + Zn^2+ --> Fe^2+ + Zn Fe is going from +0 to +2. Zn is going from +2 to +0. So the a million/2-reactions are: Fe ---> Fe++ + 2e- Zn++ + 2e- ---> Zn 3. Ox.+0.. . . . +2. . . . . . . .+3. . . . . +0 . . .2Al + 3Fe^+2 --> 2Al^+3 + 3Fe Al is going from +0 to +3. Fe is going from +2 to +0. So the a million/2-reactions are: 2(Al ---> Al+++ + 3e-) = 2Al ---> 2Al+++ + 6e- 3(Fe++ + 2e- ---> Fe) = 3Fe++ + 6e- ---> 3Fe