AC current is produced by which of the following things?
2 answers:
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Answer:
The balanced equation for this reaction will be
→
We can see that 1 mole of methane requires 4 moles of fluorine but we have 0.41 moles of CH4 and 0.56mole of F2
So using the unitary method we will get that
- 1 mole of CH4 → 4 mole of 4 mole of fluorine
- 0.41 mole of methane → 4*0.41 = 1.64 mole of fluorine for complete reaction
but we have only 0.56 mole of fluorine that means fluorine is the limiting reagent and the product will only be formed by only this amount of fluorine.
- 4 moles of fluorine → 1 mole of CF4
- 0.56 mole →
= 0.14mole of CF4
- 4 moles of fluorine → 4 moles of HF
- 0.56 mole of fluorine → 0.56 mole of HF
now to find the heat released we have the formula as
DELTA H = n * Delta H of product - n *delta H of reactant
where n is the moles of the reactant and product.
note: since no information is given about the enthalpies of the species we leave it on general equation also you need to add the product side enthalpy of the species present and similarly on the product side.
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