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3 years ago

Why can pure metals conduct electricity?

Chemistry

2 answers:

Romashka-Z-Leto [24]3 years ago

7 0

Answer:

The structure and bonding of metals explains their properties:

They are electrical conductors because their delocalised electrons carry electrical charge through the metal.

They are good conductors of thermal energy because their delocalised electrons transfer energy.

They have high melting points and boiling points, because the metallic bonding in the giant structure of a metal is very strong - large amounts of energy are needed to overcome the metallic bonds in melting and boiling.

They are malleable, which means they can be bent and shaped easily. In pure metals, the atoms are arranged in neat layers, and when a force is applied to the metal (eg by being hit with a hammer), the layers of metal atoms can slide over each other, giving the metal a new shape.

MaRussiya [10]3 years ago

4 0

Answer:

they have delocolised electrons which can carry electrical charges through the metals.

Explanation:

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Answer:

Energy transfer is the movement of energy from one place to another.

Explanation:

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2 years ago

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Suppose that you use a pair of chopsticks and apply a force of 1 N over a distance of 0.01 m. How much work do you do? If the ou

Leto [7]

Explanation:

Work done = force * perpendicular distance

= 1 * 0.01 = 0.01 joules

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3 years ago

Atoms of element X weigh 32 times more than atoms of element Y. A compound has the formula: XY2 The ratio of the mass of X to th

tamaranim1 [39]

Answer:

16:1

Explanation:

Atoms of element X weigh 32 times more than atoms of element Y. We can write this in a symbolic way.

mX = 32 mY [1]

where,

mX and mY are the masses of X and Y, respectively

A compound has the formula: XY₂, that is, in 1 molecule of XY₂ there is 1 atom of X and 2 atoms of Y. The ratio of the mass of X to the mass of Y in this compound equals:

mX/2 mY [2]

If we substitute [1] in [2], we get:

mX/2 mY = 32 mY/2 mY = 16 = 16:1

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3 years ago

Combine the two half-reactions that give the spontaneous cell reaction with the smallest E∘. Fe2+(aq)+2e−→Fe(s) E∘=−0.45V I2(s)+

Iteru [2.4K]

Answer: The spontaneous cell reaction having smallest $E^o$ is $I_2+Cu\rightarrow Cu^{2+}+2I^-$

Explanation:

We are given:

$E^o_{(Fe^{2+}/Fe)}=-0.45V\\E^o_{(I_2/I^-)}=0.54V\\E^o_{(Cu^{2+}/Cu)}=0.34V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, iodine will always undergo reduction reaction, then copper and then iron.

The equation used to calculate electrode potential of the cell is:

$E^o_{cell}=E^o_{oxidation}+E^o_{reduction}$

The combination of the cell reactions follows:

Case 1:

Here, iodine is getting reduced and iron is getting oxidized.

The cell equation follows:

$I_2(s)+Fe(s)\rightarrow Fe^{2+}(aq.)+2I^-(aq.)$

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-$ $E^o_{oxidation}=0.45V$

Reduction half reaction: $I_2(s)+2e^-\rightarrow 2I_-(aq.)$ $E^o_{reduction}=0.54V$

$E^o_{cell}=0.45+0.54=0.99V$

Thus, this cell will not give the spontaneous cell reaction with smallest $E^o_{cell}$

Case 2:

Here, iodine is getting reduced and copper is getting oxidized.

The cell equation follows:

$I_2(s)+Cu(s)\rightarrow Cu^{2+}(aq.)+2I^-(aq.)$

Oxidation half reaction: $Cu(s)\rightarrow Cu^{2+}(aq.)+2e^-$ $E^o_{oxidation}=-0.34V$

Reduction half reaction: $I_2(s)+2e^-\rightarrow 2I_-(aq.)$ $E^o_{reduction}=0.54V$

$E^o_{cell}=-0.34+0.54=0.20V$

Thus, this cell will give the spontaneous cell reaction with smallest $E^o_{cell}$

Case 3:

Here, copper is getting reduced and iron is getting oxidized.

The cell equation follows:

$Cu^{2+}(aq.)+Fe(s)\rightarrow Fe^{2+}(aq.)+Cu(s)$

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-$ $E^o_{oxidation}=0.45V$

Reduction half reaction: $Cu^{2+}(aq.)+2e^-\rightarrow Cu(s)$ $E^o_{reduction}=0.34V$

$E^o_{cell}=0.45+0.34=0.79V$

Thus, this cell will not give the spontaneous cell reaction with smallest $E^o_{cell}$

Hence, the spontaneous cell reaction having smallest $E^o$ is $I_2+Cu\rightarrow Cu^{2+}+2I^-$

7 0

3 years ago