What are the chemical equations and net ionic equations for the following 6 reac

Question

What are the chemical equations and net ionic equations for the following 6 reactions:

KMnO4 + HsSO4 + FeSO4 -->

KMnO4 + NaOH + K2C2O4 -->

KMnO4 + H2SO4 + K2C2O4 -->

K2Cr2O7 + H2SO4 + NaNO2 -->

K2Cr2O7 + H2SO4 + FeSO4 -->

KI + HNO3 -->

What are the chemical equations and net ionic equations for the following 6 reactions: KMnO4 + HsSO4 + FeSO4  KMnO4 + NaOH + K2C2O4 Rightarrow KMnO4 + H2SO4 + K2C2O4 Rightarrow K2Cr2O7 + H2SO4 + NaNO2 Rightarrow K2Cr2O7 + H2SO4 + FeSO4 Rightarrow KI + HNO3 Rightarrow

Explanation / Answer

1)

KMnO4 + FeSO4 + H2SO4 ? Fe2(SO4)3 + K2SO4 + MnSO4 + H2O

Mn goes from +7 to +2, => -5
Fe goes from +2 to +3, => +1

You need 5 Fe atoms per Mn atom.

KMnO4 + 5FeSO4 + ?H2SO4 ? 2.5 Fe2(SO4)3 + 0.5K2SO4 + MnSO4 + ?H2O

Since fractions look silly in equations, multiply all by 2.

2KMnO4 + 10FeSO4 + ?H2SO4 ? 5Fe2(SO4)3 + K2SO4 + 2MnSO4 + ?H2O

Now balance the rest.

2KMnO4 + 10FeSO4 + 8H2SO4 ? 5Fe2(SO4)3 + K2SO4 + 2MnSO4 + 8H2O

Fe++ -e = Fe+++
MnO4- + 8 H+ + 5 e = Mn++ + 4HOH
--------------------------------------...
5 Fe++ + MnO4- + 8H+ = 5Fe+3 + Mn++ + 4HOH

10FeSO4 + 2KMnO4 + 8H2SO4 = 5Fe2(SO4)3 + 2MnSO4 + K2SO4 + 8HOH

2)

2MnO4- + 5H2C2O4 + 6H+ -------------> 2Mn2+ + 8H2O + 10CO2

3)

The balanced equation is:

5K2C2O4 +8 H2SO4 + 2KMnO4 --> 2 MnSO4 + 10CO2 + 8H2O + 6K2SO4

MnO4- + 8H+ + 5e- ---> Mn2+ + 4H2O
and C2O42- ---> 2CO2 + 2e-

It is then very simple as you multiply the oxalate equation by 5, permanganate eqtn by 2 [to get 10 electrons in each which are transferred, which will then cancel on adding]

this gives
5C2O4^2-(aq) + 2MnO4^-(aq) + 16H^+(aq) -->
2Mn^2+(aq) + 10CO2(g) + 8H2O(l)

reducing agent : Oxalate oxidizing agent : Permanganate

4)

5 K2Cr2O7 + 35 NaNO2 + 20 H2SO4 = 35 NaNO3 + 5 Cr2(SO4)3 + 5 K2SO4 + 2 H20

5)

Fe is +2 in FeSO4 and goes to +3 in Fe2(SO4)3. So Fe loses 1e-, and in 2FeSO4 ===> Fe2SO4)3 two Fe's lose 2e-.

Cr is +6 in K2Cr2O7 and goes to +3 in Cr2(SO4)3. So each Cr gains 3e-, and two Cr's gain 6e-.

Electrons lost equal electrons gained, so multiply the Fe compounds by 3:

6FeSO4 + K2Cr2O7 + H2SO4 ===> 3Fe2(SO4)3 + Cr2(SO4)3 + K2SO4 + H2O

You have 2 K's on the left and right, so that means only one K2SO4. That gives you 13 SO4's on the right and only 6 on the left. So you must have 7H2SO4. That gives you 14 H's on the left, so you must have 7H2O on the right.

6FeSO4 + K2Cr2O7 + 7H2SO4 ===> 3Fe2(SO4)3 + Cr2(SO4)3 + K2SO4 + 7H2O

K2Cr2O7 + FeSO4 + H2SO4----> Fe2(SO4)3+ K2SO4+Cr2(SO4)3+ H2O
so oxidation half ionic equation ---------
Fe+2 --------------> Fe+3 + e-
reduction half ionic equation---------
Cr2O7-2 + 14H+ + 6e- --------------> 2Cr+ 3 + 7H2O
balanced full equation ---------
6Fe+2 + Cr2O-2 +14 H+ ------------> 6Fe+3 + 2Cr+3 + 7 H2O

7)

it's a redox reaction:
N+5 in nitric acid is reduced to N+4 in NO2 (gains 1 electron)
I- in potassium iodide is oxidized to I0 in I2 (loses 1 electron)

Since both species lose and gain the same number of electrons, let their coefficients be the same: 1.
Now to the adjustments:
Due to the presence of I2 in the products, the coefficient of KI should be 2, which directly implies on KNO3 coeff. being 2:

2KI + HNO3 -> 2KNO3 + I2 + NO2 + H2O

Since we need two molecules of nitric acid to oxidize two KI, two molecules of nitric acid will be reduced, whilst the other two won't (will be part of potassium nitrate), so HNO3 coeff. is 4:

2KI + 4HNO3 -> 2KNO3 + I2 + NO2 + H2O

So the coefficient of NO2 is 2(so there are 4 N atoms in both sides),

2KI + 4HNO3 -> 2KNO3 + I2 + 2NO2 + H2O

And two molecules of water! (to make it match)

2KI + 4HNO3 -> 2KNO3 + I2 + 2NO2 + 2H2O

balanced!

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