495 cm3 of oxygen gas and 877 cm3 of nitrogen gas, both at 25.0 C and 114.7 kpa, are injected into an evacuated 536 cm3 flask. F
2 answers:
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Answer :
(a) The initial cell potential is, 0.53 V
(b) The cell potential when the concentration of has fallen to 0.500 M is, 0.52 V
(c) The concentrations of and
when the cell potential falls to 0.45 V are, 0.01 M and 1.59 M
Explanation :
The values of standard reduction electrode potential of the cell are:
From this we conclude that, the zinc (Zn) undergoes oxidation by loss of electrons and thus act as anode. Nickel (Ni) undergoes reduction by gain of electrons and thus act as cathode.
The half reaction will be:
Reaction at anode (oxidation) :
Reaction at cathode (reduction) :
The balanced cell reaction will be,
First we have to calculate the standard electrode potential of the cell.
(a) Now we have to calculate the cell potential.
Using Nernest equation :
where,
n = number of electrons in oxidation-reduction reaction = 2
= emf of the cell = ?
Now put all the given values in the above equation, we get:
(b) Now we have to calculate the cell potential when the concentration of has fallen to 0.500 M.
New concentration of = 1.50 - x = 0.500
x = 1 M
New concentration of = 0.100 + x = 0.100 + 1 = 1.1 M
Using Nernest equation :
Now put all the given values in the above equation, we get:
(c) Now we have to calculate the concentrations of and
when the cell potential falls to 0.45 V.
Using Nernest equation :
Now put all the given values in the above equation, we get:
The concentration of = 1.50 - x = 1.50 - 1.49 = 0.01 M
The concentration of = 0.100 + x = 0.100 + 1.49 = 1.59 M
Answer:
B) 0.32 %
Explanation:
Given that:
Concentration = 1.8 M
Considering the ICE table for the dissociation of acid as:-
The expression for dissociation constant of acid is:
Solving for x, we get:
<u>x = 0.00568 M</u>
Percentage ionization =
<u>Option B is correct.</u>