Question

In: Chemistry

How to balance redox reactions in acid? a. Br2 + NO2^- ---> Br^- + NO3^- b....

How to balance redox reactions in acid?

a. Br2 + NO2^- ---> Br^- + NO3^-

b. H2O2 + Cr^+3 ---> Cr2O7^-2 + H2O

c. Cr2O7^-2 + CH3OH ---> Cr+3 + CH3COOH

d. NO2^- + I^- --> N2 + I2

I believe a. is NO2^-1 + H2O + 2Br2 --> 2Br^- + NO3^- + 2H+ but am very confused about the rest

Solutions

Expert Solution

a. split:

Br2 = Br-

NO2- = NO3-

balance Br

Br2 = 2Br-

NO2- = NO3-

balance O

Br2 = 2Br-

H2O + NO2- = NO3-

balance H

Br2 = 2Br-

H2O + NO2- = NO3- + 2H+

balance e-

2e- + Br2 = 2Br-

H2O + NO2- = NO3- + 2H+ +2e-

add

2e- + Br2 + H2O + NO2- = NO3- + 2H+ +2e- + 2Br-

Br2 + H2O + NO2- = NO3- + 2H+ +2Br-

b.

H2O2 = H2O

Cr+3 = Cr2O7-2

--

H2O2 = H2O + H2O

7H2O + 2Cr+3 = Cr2O7-2

--

2e- + 2H+ + H2O2 = 2H2O

7H2O + 2Cr+3 = Cr2O7-2 + 14H+ + 6e-

--

6e- + 6H+ + 3H2O2 =6H2O

7H2O + 2Cr+3 = Cr2O7-2 + 14H+ + 6e-

--

6e- + 6H+ + 3H2O2 + 7H2O + 2Cr+3 = Cr2O7-2 + 14H+ + 6e- + 6H2O

--

3H2O2 + 7H2O + 2Cr+3 = Cr2O7-2 + 8H+ +6H2O

c.

CrO7-2 + CH3OH = Cr+3 + CH3COO-

CH3OH = CH3COO-

Cr2O7-2 = 2Cr+3

--

H2O + CH3OH = CH3COO-

Cr2O7-2 = 2Cr+3 + 7H2O

--

H2O + CH3OH = CH3COO- + 3H+

14H+ + Cr2O7-2 = 2Cr+3 + 7H2O

--

H2O + CH3OH = CH3COO- + 3H+ + 2e-

6e- + 14H+ + Cr2O7-2 = 2Cr+3 + 7H2O

--

3H2O + 3CH3OH = 3CH3COO- + 9H+ + 6e-

6e- + 14H+ + Cr2O7-2 = 2Cr+3 + 7H2O

--

6e- + 14H+ + Cr2O7-2 + 3H2O + 3CH3OH = 3CH3COO- + 9H+ + 6e- + 2Cr+3 + 7H2O

sikmplify

14H+ + Cr2O7-2 + 3CH3OH = 3CH3COOH + 6H+ 2Cr+3 + 4H2O

--

8H+ + Cr2O7-2 + 3CH3OH = 3CH3COOH + 2Cr+3 + 4H2O

d.

NO2- = N2

I- = I2

--

2NO2- = N2 + 4H2O

2I- = I2 + 2e-

--

6e- + 8H+ + 2NO2- = N2 + 4H2O

2I- = I2 + 2e-

---

6e- + 8H+ + 2NO2- = N2 + 4H2O

6I- = 3I2 + 6e-

add

6I- + 6e- + 8H+ + 2NO2- = N2 + 4H2O + 3I2 + 6e-

6I-  + 8H+ + 2NO2- = N2 + 4H2O + 3I2

queen_honey_blossom answered 2 years ago
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