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In-s [12.5K]
3 years ago
15

Using the following data, determine the standard cell potential E^o cell for the electrochemical cell constructed using the foll

owing reaction, where zinc is the anode and lead is the cathode.
Zn(s) + Pb2+(aq) -> Zn2+(aq) + Pb(s)

Half-reaction: Standard Reduction Potential:

Zn2+(aq) + 2e- -> Zn(s)= -0.763

Pb2+(aq) + 2e- -> Pb(s)= -0.126

a. -0.889 V

b. +0.889 V

c. +0.637 V

d. +1.274 V

e. -0.637 V
Chemistry
1 answer:
Nana76 [90]3 years ago
8 0

Answer: C)  0.637 V

Explanation:

The balanced redox reaction is:

Zn(s)+Pb^{2+}(aq)\rightarrow Zn^{2+}(aq)+Pb(s)

Here Zn undergoes oxidation by loss of electrons, thus act as anode. Lead undergoes reduction by gain of electrons and thus act as cathode.

E^0=E^0_{cathode}- E^0_{anode}

Where both E^0 are standard reduction potentials.

Zn^{2+}(aq)+2e^-\rightarrow Zn(s)= -0.763

Pb^{2+}(aq)+2e^-\rightarrow Pb(s)= -0.126

E^0_{[Zn^{2+}/Zn]}=-0.763V

E^0_{[Pb^{2+}/Pb]}=-0.126V

E^0=E^0_{[Pb^{2+}/Pb]}- E^0_{[Zn^{2+}/Zn]}

E^0=-0.126-(-0.763V)=0.637V

The standard emf of a cell is 0.637 V

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3 years ago
combustion analysis of a hydrocarbon produced 33.01g CO2 and 13.51g H2O. Calculate the empirical formula for the hydrocarbon
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Answer:

\rm CH_2.

Explanation:

Carbon and hydrogen are the only two elements in a hydrocarbon. When a hydrocarbon combusts completely in excess oxygen, the products would be \rm CO_2 and \rm H_2O. The \rm C and \rm H would come from the hydrocarbon, while the \rm O atoms would come from oxygen.

Look up the relative atomic mass of these three elements on a modern periodic table:

  • \rm C: 12.011.
  • \rm H: 1.008.
  • \rm O: \rm 15.999.

Calculate the molar mass of \rm CO_2 and \rm H_2O:

M(\mathrm{CO_2}) = 12.011 + 2 \times 15.999 = 44.009\; \rm g \cdot mol^{-1}.

M(\mathrm{H_2O}) = 2 \times 1.008 + 15.999 = 18.015\; \rm g \cdot mol^{-1}

Calculate the number of moles of \rm CO_2 molecules in 33.01\; \rm g of \rm CO_2\!:

\displaystyle n(\mathrm{CO_2}) = \frac{m(\mathrm{CO_2})}{M(\mathrm{CO2})} = \frac{33.01\; \rm g}{44.009\; \rm g\cdot mol^{-1}} \approx 0.7501\; \rm mol.

Similarly, calculate the number of moles of \rm H_2O molecules in 13.51\; \rm g of \rm H_2O\!:

\displaystyle n(\mathrm{H_2O}) = \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} = \frac{13.51\; \rm g}{18.015\; \rm g\cdot mol^{-1}} \approx 0.7499\; \rm mol.

Note that there is one carbon atom in every \rm CO_2 molecule. Approximately0.7501\; \rm mol of \rm CO_2\! molecules would correspond to the same number of \rm C atoms. That is: n(\mathrm{C}) \approx 0.7501\; \rm mol.

On the other hand, there are two hydrogen atoms in every \rm H_2O molecule. approximately 0.7499\; \rm mol of \rm H_2O molecules would correspond to twice as many \rm H\! atoms. That is: n(\mathrm{H}) \approx 2 \times 0.7499 \; \rm mol\approx 1.500\; \rm mol.

The ratio between the two is: n(\mathrm{C}): n(\mathrm{H}) \approx 1:2.

The empirical formula of a compound gives the smallest whole-number ratio between the elements. For this hydrocarbon, the empirical formula would be \rm CH_2.

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