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Ivenika [448]
3 years ago
14

How many moles of water are produced from 0.500 moles of oxygen gas?

Chemistry
1 answer:
solmaris [256]3 years ago
6 0
As far as I know NONE.....
Explanation:
Sodium peroxide can be thermolyzed to give dioxygen gas...
N
a
2
O
2
(
s
)
+
Δ
→
N
a
2
O
(
s
)
+
1
2
O
2
(
g
)
↑
⏐
⏐
⏐

But with water, we simply get an acid base reaction....
N
a
2
O
2
(
s
)
+
2
H
2
O
(
l
)
→
2
N
a
O
H
(
a
q
)
+
H
2
O
2
(
a
q
)
...
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<u>Answer:</u> The concentration of Au^{3+} in the solution is 1.87\times 10^{-14}M

<u>Explanation:</u>

The given cell is:

Pt(s)|H_2(g.1atm)|H^+(aq.,1.0M)||Au^{3+}(aq,?M)|Au(s)

Half reactions for the given cell follows:

<u>Oxidation half reaction:</u> H_2(g)\rightarrow 2H^{+}(1.0M)+2e^-;E^o_{H^+/H_2}=0V ( × 3)

<u>Reduction half reaction:</u> Au^{3+}(?M)+3e^-\rightarrow Au(s);E^o_{Au^{3+}/Au}=1.50V ( × 2)

<u>Net reaction:</u> 3H_2(s)+2Au^{3+}(?M)\rightarrow 6H^{+}(1.0M)+2Au(s)

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

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E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[H^{+}]^6}{[Au^{3+}]^2}

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E^o_{cell} = standard electrode potential of the cell = +1.50 V

n = number of electrons exchanged = 6

[Au^{3+}]=?M

[H^{+}]=1.0M

Putting values in above equation, we get:

1.23=1.50-\frac{0.059}{6}\times \log(\frac{(1.0)^6}{[Au^{3+}]^2})

[Au^{3+}]=1.87\times 10^{-14}M

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