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madreJ [45]
4 years ago
10

A galvanic cell is powered by the following redox reaction:2Fe+3(aq) + H2 (g) + 2OH−(aq) → 2Fe+2 (aq) + 2H2O (l)Answer the follo

wing questions about this cell. If you need any electrochemical data, be sure you get it from the ALEKS Data tab.Write a balanced equation for the half-reaction that takes place at the cathode. Write a balanced equation for the half-reaction that takes place at the anode. Calculate the cell voltage under standard conditions. =E0V
Chemistry
1 answer:
ICE Princess25 [194]4 years ago
7 0

Answer:

Anode: H₂(g) + 2 OH⁻(aq) → 2 H₂O(l) + 2 e⁻

Cathode: 2 Fe⁺³(aq) + 2 e⁻ → 2 Fe⁺²(aq)

E° = 1.60 V

Explanation:

Let's consider the reaction taking place in a galvanic cell.

2 Fe⁺³(aq) + H₂(g) + 2 OH⁻(aq) → 2 Fe⁺²(aq) + 2 H₂O(l)

The corresponding half-reactions are:

Anode (oxidation): H₂(g) + 2 OH⁻(aq) → 2 H₂O(l) + 2 e⁻   E°red = - 0.83 V

Cathode (reduction): 2 Fe⁺³(aq) + 2 e⁻ → 2 Fe⁺²(aq)        E°red = 0.77 V

The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.

E° = E°red, cat - E°red, an = 0.77 V - (-0.83 V) = 1.60 V

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What are the empirical and molecular formulas of a hydrocarbon if combustion of 2.10 g of the compound yields 6.59 g co2 and 2.7
mariarad [96]

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    <u><em>Explanation</em></u>

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<h3><u><em> </em></u>Empirical formula  calculation</h3>

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moles =mass/molar mass

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that  is  for  CO2 = 0.15/0.15  =1

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therefore  since  there 1 atom  of C  in product side there  must be 1 atom of C  in reactant  side.

In addition  there is 2 H atom in product  side  which should be the  same  in reactant side.  

From information above the empirical formula is therefore = CH2


Molecular formula  calculation

[CH2}n= 84 g/mol

[12+ (1x2)] n = 84 g/mol

14 n =  84 g/mol

n = 6

multiply the  each subscript  in CH2  by  6

 Therefore the molecular formula = C6H12




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