Electrons always fill orbitals in the same order. Each s orbital holds 2 electrons, each set of p orbitals holds 6 electrons, ea
ch set of d orbitals holds 10 electrons, and each set of f orbitals holds 14 electrons. The order in which orbitals are filled, from first to last, is:
1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p
To what element does the following electron configuration correspond?
1s22s22p4
A. aluminum: 13 electrons
B. oxygen: 8 electrons
C. boron: 5 electrons
1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p
To what element does the following electron configuration correspond?
1s22s22p4
A. aluminum: 13 electrons
B. oxygen: 8 electrons
C. boron: 5 electrons
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