In which substance does hydrogen have an oxidation number of zero?

Answers

Answer 1
Answer: Answer is: H₂ (molecule of hydrogen).

Missing question: A) LiH B) H₂O C) H₂S D) H₂.
1) In lithium hydride (LiH), lithium has oxidation number +1, than hydrogen must have oxidation number -1, because compound has neutral charge.
2) In water molecule (H₂O), oxygen has oxidation number -2 and hydrogen has +1: 2x + (-2) = 0; x = +1.
3) In hydrogen sulfide (H₂S), sulfur has oxidation number -2 and hydrogen has +1: 2x + (-2) = 0; x = +1.
4) In molecule of hydrogen there is nonpolar covalent bond.

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77. A voltaic cell consists of a Zn/Zn2+ half-cell and a Ni/Ni2+ half-cell at 25°C. The initial concentrations of Ni2+ and Zn2+ are 1.50 M and 0.100 M, respectively. a. What is the initial cell potential? b. What is the cell potential when the concentration of Ni2+ has fallen to 0.500 M? c. What are the concentrations of Ni2+ and Zn2+ when the cell potential falls to 0.45 V?

Answers

Answer :

(a) The initial cell potential is, 0.53 V

(b) The cell potential when the concentration of Ni^(2+) has fallen to 0.500 M is, 0.52 V

(c) The concentrations of Ni^(2+) and Zn^(2+) 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:

E^0_([Ni^(2+)/Ni])=-0.23V

E^0_([Zn^(2+)/Zn])=-0.76V

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) :Zn\rightarrow Zn^(2+)+2e^-     E^0_([Zn^(2+)/Zn])=-0.76V

Reaction at cathode (reduction) :Ni^(2+)+2e^-\rightarrow Ni     E^0_([Ni^(2+)/Ni])=-0.23V

The balanced cell reaction will be,  

Zn(s)+Ni^(2+)(aq)\rightarrow Zn^(2+)(aq)+Ni(s)

First we have to calculate the standard electrode potential of the cell.

E^o=E^o_(cathode)-E^o_(anode)

E^o=E^o_([Ni^(2+)/Ni])-E^o_([Zn^(2+)/Zn])

E^o=(-0.23V)-(-0.76V)=0.53V

(a) Now we have to calculate the cell potential.

Using Nernest equation :

E_(cell)=E^o_(cell)-(0.0592)/(n)\log ([Zn^(2+)])/([Ni^(2+)])

where,

n = number of electrons in oxidation-reduction reaction = 2

E_(cell) = emf of the cell = ?

Now put all the given values in the above equation, we get:

E_(cell)=0.53-(0.0592)/(2)\log ((0.100))/((1.50))

E_(cell)=0.49V

(b) Now we have to calculate the cell potential when the concentration of Ni^(2+) has fallen to 0.500 M.

New concentration of Ni^(2+) = 1.50 - x = 0.500

x = 1 M

New concentration of Zn^(2+) = 0.100 + x = 0.100 + 1 = 1.1 M

Using Nernest equation :

E_(cell)=E^o_(cell)-(0.0592)/(n)\log ([Zn^(2+)])/([Ni^(2+)])

Now put all the given values in the above equation, we get:

E_(cell)=0.53-(0.0592)/(2)\log ((1.1))/((0.500))

E_(cell)=0.52V

(c) Now we have to calculate the concentrations of Ni^(2+) and Zn^(2+) when the cell potential falls to 0.45 V.

Using Nernest equation :

E_(cell)=E^o_(cell)-(0.0592)/(n)\log ([Zn^(2+)+x])/([Ni^(2+)-x])

Now put all the given values in the above equation, we get:

0.45=0.53-(0.0592)/(2)\log ((0.100+x))/((1.50-x))

x=1.49M

The concentration of Ni^(2+) = 1.50 - x = 1.50 - 1.49 = 0.01 M

The concentration of Zn^(2+) = 0.100 + x = 0.100 + 1.49 = 1.59 M

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Answers

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Answers

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Answers

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Answers

Answer: cash crops

Explanation:

trust

Answer: Cash Crops

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The interactive forces between particles are... A) less effective in solids than in liquids B) more effective in solids than in liquids C) equal to that of a gas D) zero

Answers

the answer is option b
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