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3 years ago

A fuel cell designed to react grain alcohol with oxygen has the following net reaction:

Chemistry

1 answer:

sertanlavr [38]3 years ago

8 0

Answer:

b. E = 2,28V

Explanation:

The maximum work is the same than ΔG. As ΔG could be written as:

ΔG = nFE (1)

Where n is moles of electrons transferred, F is faraday constant (96485 J/Vmol) and E is the voltage of the cell.

For the reaction:

CH₃OH(l) + ³/₂O₂(g) → CO₂(g) + 2H₂O(l)

The oxidation state of C in CH₃OH is -2 but in CO₂ is +4, that means transferred electrons are +4 - -2 = 6e⁻

Replacing in (1):

1320x10³ J = 6mol e⁻×96485J/Vmol×E

E = 2,28V

I hope it helps!

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Which of the following views of weather is an example of systems thinking

Luda [366]

Answer:

Explanation:

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3 years ago

Balance the following equation Sa+ F2---> SF6

deff fn [24]

Answer:

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3 years ago

Calculate E ° for the half‑reaction, AgCl ( s ) + e − − ⇀ ↽ − Ag ( s ) + Cl − ( aq ) given that the solubility product constant

antoniya [11.8K]

Answer: The value of $E^{o}$ for the half-cell reaction is 0.222 V.

Explanation:

Equation for solubility equilibrium is as follows.

$AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)$

Its solubility product will be as follows.

$K_{sp} = [Ag^{+}][Cl^{-}]$

Cell reaction for this equation is as follows.

$Ag(s)| AgCl(s)|Cl^{-}(0.1 M)|| Ag^{+}(1.0 M)| Ag(s)$

Reduction half-reaction: $Ag^{+} + 1e^{-} \rightarrow Ag(s)$ , $E^{o}_{Ag^{+}/Ag} = 0.799 V$

Oxidation half-reaction: $Ag(s) + Cl^{-}(aq) \rightarrow AgCl(s) + 1e^{-}$ , $E^{o}_{AgCl/Ag}$ = ?

Cell reaction: $Ag^{+}(aq) + Cl^{-}(aq) \rightarrow AgCl(s)$

So, for this cell reaction the number of moles of electrons transferred are n = 1.

Solubility product, $K_{sp} = [Ag^{+}][Cl^{-}]$

= $1.77 \times 10^{-10}$

Therefore, according to the Nernst equation

$E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}$

At equilibrium, $E_{cell}$ = 0.00 V

Putting the given values into the above formula as follows.

$E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}$

$0.00 = E^{o}_{cell} - \frac{0.0592 V}{1} log \frac{1}{[Ag^{+}][Cl^{-}]}$

$E^{o}_{cell} = \frac{0.0592}{1} log \frac{1}{K_{sp}}$

= $0.0591 V \times log \frac{1}{1.77 \times 10^{-10}}$

= 0.577 V

Hence, we will calculate the standard cell potential as follows.

$E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode}$

$0.577 V = E^{o}_{Ag^{+}/Ag} - E^{o}_{AgCl/Ag}$

$0.577 V = 0.799 V - E^{o}_{AgCl/Ag}$

$E^{o}_{AgCl/Ag}$ = 0.222 V

Thus, we can conclude that value of $E^{o}$ for the half-cell reaction is 0.222 V.

3 0

3 years ago

Why do group 1 cations form precipitates when mixed with hcl?

Xelga [282]

Both of you are overlooking a pretty big component of the question...the Group I cation isn't being dissociated into water. We're testing the solubility of the cation when mixed with HCl. And this IS a legitimate question, seeing as our lab manual is the one asking.

By the way, the answer you're looking for is "Because Group I cations have insoluble chlorides". 

"In order...to distinguish cation Group I, one adds HCl to a sample. If a Group I cation is present in the sample, a precipitate will form."

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3 years ago

An Arrhenius acid increases the concentration of hydrogen ions when dissolved in solution. True False

shusha [124]

Answer:

True

Explanation:

The arrhenius theory describes an acid as a substance that produces excess hydrogen ions when it interacts with water. Therefore, they increase the concentration of hydrogen ions in solutions.

An arrhenius base interacts with water to yield excess hydroxide ions in aqeous solutions.

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3 years ago