Composition, and a c<span>hemical reaction.</span>
Answer & Explanation:
(a)

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
Explanation:
From first source, kinetic energy (
) ejected is 1 eV and wavelength of light is
.
From second source, kinetic energy (
) ejected is 4 eV and wavelength of light is
.
Relation between work function, wavelength, and kinetic energy is as follows.
K.E =
where, h = Plank's constant =
J.s
c = speed of light =
m/s
Also, it is known that 1 eV =
J
Therefore, substituting the values in the above formula as follows.
=
1 eV =
........... (1)
=
........... (2)
Now, divide equation (2) by 2. Therefore, it will become
......... (3)
Now, subtract equation (3) from equation (1), we get the following.

= 
= 2 eV
Thus, we can conclude that work function of the metal is 2 eV.
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