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Elanso [62]
3 years ago
5

Where does reduction occur in a battery

Chemistry
1 answer:
morpeh [17]3 years ago
8 0
~Hello there!

Your question: Where does reduction occur in a battery?

Your answer: Reduction occurs at the Cathode in a battery.

Any questions^ ?

Happy Studying!
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Baking soda is sodium bicarbonate in anhydrous form without any water of crystallisation and it is widely used as dry fire extinguisher because of its alkali nature.
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Determine whether the following hydroxide ion concentrations ([OH−]) correspond to acidic, basic, or neutral solutions by estima
Rzqust [24]

Answer:

See explanation below

Explanation:

To do this, we will use the following expression to calculate the [H⁺]:

[H⁺] = Kw / [OH⁻]

[H⁺] is the same as [H₃O⁺]. So we have the [OH⁻] so, let's replace every value into the above expression to calculate the hydronium concentration and say if it's acidic, basic or neutral. This can be known because if the [H⁺] > 1x10⁻⁷ M the solution is acidic. If it's [H⁺] < 1x10⁻⁷ M the solution is basic, and if it's [H⁺] = 1x10⁻⁷ M the solution is neutral.

a) [H⁺] = 1x10⁻¹⁴ / 6x10⁻¹² = 1.67x10⁻³ M. Acidic.

b) [H⁺] = 1x10⁻¹⁴ / 9x10⁻⁹ = 1.11x10⁻⁶ M. Acidic.

c) [H⁺] = 1x10⁻¹⁴ / 8x10⁻¹⁰ = 1.25x10⁻⁵ M. Acidic.

d) [H⁺] = 1x10⁻¹⁴ / 7x10⁻¹³ = 0.0143 M. Acidic.

e) [H⁺] = 1x10⁻¹⁴ / 2x10⁻² = 5x10⁻¹³ M. Basic.

f) [H⁺] = 1x10⁻¹⁴ / 9x10⁻⁴ = 1.11x10⁻¹¹ M. Basic.

g) [H⁺] = 1x10⁻¹⁴ / 5x10⁻⁵ = 2x10⁻¹⁰ M. Basic.

h) [H⁺] = 1x10⁻¹⁴ / 1x10⁻⁷ = 1x10⁻⁷ M. Neutral.

Part B.

In this part, we'll use the following expression and replace the given values:

[OH⁻] = Kw / [H⁺]

Replacing the values:

[OH⁻] = 1x10⁻¹⁴ / 5.2x10⁻⁵

[OH⁻] = 1.92x10⁻¹⁰ M

PArt C:

In this case, we will use expression of part A, and replace the given values:

[H⁺] = 1x10⁻¹⁴ / 2.7x10⁻²

[H⁺] = 3.7x10⁻¹³ M

8 0
3 years ago
Tarnish on copper is the compound CuO. A tarnished copper plate is placed in an aluminum pan of boiling water. When enough salt
Jet001 [13]

Answer:  2.00 V

Explanation:

The balanced redox reaction is:

2Al+3Cu^{2+}\rightarrow 2Al^{3+}+3Cu

Here Al undergoes oxidation by loss of electrons, thus act as anode. Copper 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.

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E^0=E^0_{[Cu^{2+}/Cu]}- E^0_{[Al^{3+}/Al]}

E^0=+0.34- (-1.66V)=2.00V

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Where would electrons in the d sub level be found
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Answer:

d sub level starts from the 3rd energy level; means you can see electrons in d sub level in any energy levels after third one.

So the answer is "In any energy levels ≥ 3"

A simple information ; s sub level is in any energy level,

p sub level is in any energy level beginning from the 2nd energy level,

d sub level appears ≥ 3rd energy level

f sub level appears in 4th energy level and over.

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#MissionExam001

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