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4 years ago

What happens when a metal is reduced. Give an example of a metal being reduced

Chemistry

1 answer:

AlexFokin [52]4 years ago

6 0

Answer:

The oxidation number of the metal decreases

2 Al + Fe₂O₃ → Al₂O₃ + 2 FeO

The metal element iron, is reduced from Fe⁺³ in Fe₂O₃ to Fe⁺² in FeO

Explanation:

When an element gains electron, the element becomes reduced, hence when a metal is reduced, the metal gains electrons, which reduces the oxidation number of the metal

An example of a metal being reduced is;

2 Al + Fe₂O₃ → Al₂O₃ + 2 FeO

In the above reaction, the iron (III) oxide is reduced to iron (II) oxide by aluminium metal.

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4 years ago

A student constructs a galvanic cell that has a strip of iron metal immersed in a solution of 0.1M Fe(NO3)2 as one half-cell and

MrMuchimi

Answer:

Explanation:

The cell potential is:

ΔE°cell = E°red(red) - E°red(oxid)

Where, E°red(red) is the reduction potential of the substance that is reducing, and E°red(oxid) is the reduction potential of the substance that is oxidizing. For the reaction be spontaneous and happen, ΔE°cell > 0.

The reduction takes place in the cathode, which is the negative pole, and the oxidation in the anode, which is the positive pole. So, the electrons flow from the positive pole to the negative pole (anode to cathode).

Then, if the voltmeter measured a negative potential, it means that is was attached incorrectly. So, the anode is Fe.

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3 years ago

Which of the compounds below is not an example of a molecular solid?

Goshia [24]

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4 years ago

A critical reaction in the production of energy in biological systems is the hydrolysis of adenosine triphosphate (ATP) to adeno

Archy [21]

Answer:

ΔG° of reaction = -47.3 x $10^{3}$ J/mol

Explanation:

As we can see, we have been a particular reaction and Energy values as well.

ΔG° of reaction = -30.5 kJ/mol

Temperature = 37°C.

And we have to calculat the ΔG° of reaction in the biological cell which contains ATP, ADP and HPO4-2:

The first step is to calculate the equilibrium constant for the reaction:

Equilibrium Constant K = $\frac{[HPO4-2] x [ADP]}{ATP}$

And we have values given for these quantities in the biological cell:

[HP04-2] = 2.1 x $10^{-3}$ M

[ATP] = 1.2 x $10^{-2}$ M

[ADP] = 8.4 x $10^{-3}$ M

Let's plug in these values in the above equation for equilibrium constant:

K = $\frac{[2.1x10^{-3}] x [8.4x10^{-3}] }{[1.2 x 10^{-2}] }$

K = 1.47 x $10^{-3}$ M

Now, we have to calculate the ΔG° of reaction for the biological cell:

But first we have to convert the temperature in Kelvin scale.

Temp = 37°C

Temp = 37 + 273

Temp = 310 K

ΔG° of reaction = (-30.5 $10^{3}$ ) + (8.314)x (310K)xln(0.00147)

Where 8.314 = value of Gas Constant

ΔG° of reaction = (-30.5 x $10^{3}$ ) + (-16810.68)

ΔG° of reaction = -47.3 x $10^{3}$ J/mol

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3 years ago