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scoray [572]
4 years ago
12

What is the relationship between an alkali metals location in group 1a and its reactivity?

Chemistry
1 answer:
Helga [31]4 years ago
3 0
The are many different trends in the periodic table. The reactivity of alkali metals increases from top to bottom in Group 1A. This is the opposite of electronegativity since it decreases from top to bottom in a group due to the increase in the atomic radius of elements. From top to bottom within a group, there is also a trend which is that atomic radius increases. In the other hand, ionization and electronegativity increase from bottom to top within a group and lastly, atomic radius increases from bottom to top and from left to right.
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Consider the following unbalanced redox reactions. In each case, separate the whole reactions into half-reactions, balance the h
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Answer & Explanation:

(a)

Fe^{2+} +NO_{3}^{-}  => Fe{(OH)}_3 + N_2

reducing agent = Fe²⁺

Oxidizing agent = NO₃⁻

oxidation

Fe²⁺  ⇒ Fe(OH)₃

reduction

NO₃⁻  ⇒ N₂

Oxidation Half Reaction

(<em>redox reactions are balanced by adding appropriate H⁺ and H₂O atoms)</em>

Fe²⁺ ⇒ Fe(OH)₃

Balance O atoms

Fe²⁺ + 3H₂O ⇒ Fe(OH)₃

Balance H atoms

Fe² + 3H₂0 ⇒ Fe(OH)₃ + 3H⁺

balance Charge

Fe² + 3H₂0 ⇒ Fe(OH)₃ + 3H⁺ + e⁻..............(1)

reduction Half Reaction

NO₃⁻ ⇒ N₂

Balance N atoms

2NO₃⁻ ⇒N₂

Balance O atoms by adding appropriate H₂O

2NO₃⁻ ⇒ N₂ + 6H₂O

Balance H atoms

2NO₃⁻ + 12H⁺ ⇒ N₂ + 6H₂O

Balance Charge

2NO₃⁻ + 12H⁺ + 10e⁻⇒ N₂ + 6H₂O.................(2)

Combine Equation (1) and (2)

(1) × 10: 10Fe² + 30H₂0 ⇒ 10Fe(OH)₃ + 30H⁺ + 10e⁻

(2) × 1:   2NO₃⁻ + 12H⁺ + 10e⁻⇒ N₂ + 6H₂O

(1) + (2): 10Fe² + <u><em>30H₂0</em></u> + 2NO₃⁻ + <u><em>12H⁺</em></u> + <u><em>10e⁻</em></u> ⇒10Fe(OH)₃ + <u><em>30H⁺</em></u><u><em> </em></u>+ <em><u>10e⁻</u></em> +              

            N₂ + <u><em>6H₂O</em></u>

            10Fe² + 24H₂0 + 2NO₃⁻  ⇒ 10Fe(OH)₃ + 18H⁺ + N₂

this is the balanced reaction

REDUCTION POTENTIAL

10Fe²⁺(aq) + 10e⁻ ⇒ 10Fe(OH)₃(aq)             E°ox = 10(-0.44) = -4.4V

2NO₃⁻(aq) -  2e⁻ ⇌ N₂(g) + 18H⁺       E°red = 2(+0.80) = +1.6

10Fe² + 24H₂0 + 2NO₃⁻  ⇒ 10Fe(OH)₃ + 18H⁺ + N₂    E°cell = -2.8V

E°cell = E°red + E°ox

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