Question

In the following reactions, which substance is oxidized? Which is the oxidizing agent?

a. 2 NHO3+ SO2→H2 SO4+2 NO2 2

b. CrO3+6 HI→Cr2 O3+3 I2+3 H2 O

c. 2 MnO2+NO2-+2H+→Mn2O2+NO3-

d. 3 MnO2+2NH4+→3 Mn2++N2+6 H2O

Answer 1

Oxidation ---> to lose electrons (+1 --> +2)

Reduction ---> to gain electrons (+5 --> +3)

Oxidation agent, the one that gets reduced (gains electrons)

Reducing agent, the on that gets oxidized (loses electrons)

Now

A) 2NHO3 + SO2→H2SO4 + 2(NO2-2)

Suphur:

SO2 --> x + 2(-2) = 0

therefore oxidation number of S = +4

in the product

H2SO4

SO4 = -2 charged

SO4 = x + 4*(-2) = x-8 = -2

x = +6

S = +4 to S = +6 ... there is a loss of 2 electrons, therefore S is oxidizing!

H rarely reduces/oxidizes so lets analyze N

for N

NHO3 --> 0 charge

x + 1 + 3(-2) = x + 1-6 = 0

x = +5

IN the NO2-2

charge = -2

x + 2*(-2) = -2

x -4 = -2

x = +2

Therefore:

N goes from x = +5 to x = +2, it is gaining electrons, therefore reducing. If it reduces, it is the oxidizing agent!

B)

CrO3+6HI→Cr2O3+ 4I2+ 3H2O

Lets analyz Cr, which is comonly reduced/oxidized

CrO3 = 0

x + 2(-3) = 0

x = +6

Cr = +6

to Cr2O3 =0

2*x + 2(-3) =0

x = +3

Cr = +3

Therefore

Cr goes from +6 to +3, it is gainig +3 electrons, it is reducing! this is the oxidizing agent!

Now analyze I

HI = 0

+1 + x = 0

x = -1

In the I2 form:

I2 = 0

therefore

I = 0

it is losing -1 to 0, 1 electron, this is oxidizing, therefore the reducing agent