Answer:
8.34
Explanation:
1) how much moles of NH₃ are in the reaction;
2) how much moles of H₂ are in the reaction;
3) the required mass of the H₂.
all the details are in the attachment; the answer is marked with red colour.
Note1: M(NH₃) - molar mass of the NH₃, constant; M(H₂) - the molar mass of the H₂, constant; ν(NH₃) - quantity of NH₃; ν(H₂) - quantity of H₂.
Note2: the suggested solution is not the shortest one.
Answer : The cell potential for this reaction is 0.50 V
Explanation :
The given cell reactions is:
The half-cell reactions are:
Oxidation half reaction (anode):
Reduction half reaction (cathode):
First we have to calculate the cell potential for this reaction.
Using Nernest equation :
where,
F = Faraday constant = 96500 C
R = gas constant = 8.314 J/mol.K
T = room temperature =
n = number of electrons in oxidation-reduction reaction = 2
= standard electrode potential of the cell = +0.63 V
= cell potential for the reaction = ?
= 3.5 M
=
Now put all the given values in the above equation, we get:
Therefore, the cell potential for this reaction is 0.50 V