When the same species undergoes both oxidation and reduction in a single redox reaction, this is referred to as a disproportionation. Therefore, divide it into two equal reactions.
NO2→NO^−3
NO2→NO
and do the usual changes
First, balance the two half reactions:
3. NO2 +H2O →NO^−3 + 2 H^+ + e−
4. NO2 +2 H^+ + 2e− → NO + H2O
Now multiply one or both half-reactions to ensure that each has the same number of electrons. Here, Eqn (3) x 2 results in each half-reaction having two electrons:
5. 2 NO2 + 2 H2O → 2 NO^−3 + 4H^+ + 2e−
Now add Eqn 4 and 5 (the electrons now cancel each other):
3NO2 + 2H^+ + 2H2O → NO + 2 NO−3 + H2O + 4H+
and cancel terms that’s common to both sides:
3NO2 + H2O → NO + 2NO^−3 + 2H+
This is the net ionic equation describing the oxidation of NO2 to NO3 in basic solution.
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Answer:
Water has the greatest ΔEN
ΔEN H₂O → 3.4 - 2.1 = 1.3 Option D.
Explanation:
We should find the Electronegativity data in the Periodic table for all the elements:
C : 2.6
O: 3.4
H: 2.1
S: 2.6
N: 3.0
a. ΔEN CO₂ → 3.4 - 2.6 = 0.4
b. ΔEN H₂S → 2.6 - 2.1 = 0.5
c. ΔEN NH₃ → 3 - 2.1= 0.9
d. ΔEN H₂O → 3.4 - 2.1 = 1.3
