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3 years ago

16. When a reaction is at equilibrium

Chemistry

1 answer:

Ronch [10]3 years ago

8 0

Answer:

The forward reaction rate is equal to the reverse reaction rate

Explanation:

In many cases, a lot of people might think that equilibrium is a state when a reaction stops. This is not the case, however. Notice that equilibrium reactions are represented by double arrows indicating that two reactions take place simultaneously: both a forward and a reverse reaction.

When equilibrium is reached, the rate of a forward reaction becomes equal to the rate of a reverse reaction. The two reactions don't stop and still take place, just at equal rates. This is the reason why the concentrations of reactants and products are kept constant at equilibrium: the rates of the two reactions are balanced.

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An electrochemical cell at 25°C is composed of pure copper and pure lead solutions immersed in their respective ionis. For a 0.6

ExtremeBDS [4]

Answer :

(a) The concentration of $Pb^{2+}$ is, 0.0337 M

(b) The concentration of $Pb^{2+}$ is, $6.093\times 10^{32}M$

Solution :

<u>(a) As per question, lead is oxidized and copper is reduced.</u>

The oxidation-reduction half cell reaction will be,

Oxidation half reaction: $Pb\rightarrow Pb^{2+}+2e^-$

Reduction half reaction: $Cu^{2+}+2e^-\rightarrow Cu$

The balanced cell reaction will be,

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

Here lead (Pb) undergoes oxidation by loss of electrons, thus act as anode. Copper (Cu) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

$E^o_{[Pb^{2+}/Pb]}=-0.13V$

$E^o_{[Cu^{2+}/Cu]}=+0.34V$

$E^o=E^o_{[Cu^{2+}/Cu]}-E^o_{[Pb^{2+}/Pb]}$

$E^o=0.34V-(-0.13V)=0.47V$

Now we have to calculate the concentration of $Pb^{2+}$ .

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Pb^{2+}]}{[Cu^{2+}]}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = 0.507 V

Now put all the given values in the above equation, we get:

$0.507=0.47-\frac{0.0592}{2}\log \frac{[Pb^{2+}]}{(0.6)}$

$[Pb^{2+}]=0.0337M$

Therefore, the concentration of $Pb^{2+}$ is, 0.0337 M

<u>(b) As per question, lead is reduced and copper is oxidized.</u>

The oxidation-reduction half cell reaction will be,

Oxidation half reaction: $Cu\rightarrow Cu^{2+}+2e^-$

Reduction half reaction: $Pb^{2+}+2e^-\rightarrow Pb$

The balanced cell reaction will be,

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

Here Copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

$E^o_{[Pb^{2+}/Pb]}=-0.13V$

$E^o_{[Cu^{2+}/Cu]}=+0.34V$

$E^o=E^o_{[Pb^{2+}/Pb]}-E^o_{[Cu^{2+}/Cu]}$

$E^o=-0.13V-(0.34V)=-0.47V$

Now we have to calculate the concentration of $Pb^{2+}$ .

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cu^{2+}]}{[Pb^{2+}]}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = 0.507 V

Now put all the given values in the above equation, we get:

$0.507=-0.47-\frac{0.0592}{2}\log \frac{(0.6)}{[Pb^{2+}]}$

$[Pb^{2+}]=6.093\times 10^{32}M$

Therefore, the concentration of $Pb^{2+}$ is, $6.093\times 10^{32}M$

6 0

3 years ago

Compared to other stars we observe, the Sun appears bigger and brighter because:

Kipish [7]

Answer:

Compared to other stars we observe, the Sun appears bigger and brighter because it is much closer to earth.

Objects closer to earth appear to be much bigger and brighter.

Let me know if this helps!

3 0

3 years ago

Read 2 more answers

If the pressure changes from 1 atm to 3 atm, what does the volume of 30 L change to

DiKsa [7]

Answer: 10L

Explanation:

Given that:

Initial pressure P1 = 1 atm

New pressure P2 = 3 atm

Initial volume V2 = 30 L

New volume V2 = ?

Since pressure and volume are involved, apply the formula for Boyle's law

P1V1 = P2V2

1 atm x 30L = 3 atm x V2

30 atm L = 3 atm x V2

V2 = (30 atm L / 3 atm)

V2 = 10L

Thus, volume changed to 10 liters

5 0

3 years ago

A black hole can be considered a star that has

lubasha [3.4K]

The first option, collapsed in on itself. The star's core mass becomes so dense that the resulting gravity implodes the star. Interesting enough, the third option is kindof true too...some large and tenacious black holes that absorb other stars will form incredibly bright accretion disks around their perimeter before filling absorbing the star.

7 0

3 years ago

What occurs when the nuclei of two very small atoms combine into one large nucleus?

Svetlanka [38]

Answer:

The combination of nuclei to form a bigger and heavier nucleus is known as Fussion. The consequence of fusion is the absorption or release of energy.

7 0

3 years ago