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A standard galvanic cell is constructed so that the overall cell reaction is as given below where M is an unknown metal. 2 Al3+(

jasenka [17]

Answer:

magnesium metal

Explanation:

According to the redox reaction equation, six electrons were transferred hence n=6 and F= Faraday's constant 96500C. ∆G° is given hence E°cell can easily be calculated as follows:

From ∆G°= -nFE°cell

E°cell= -∆G°/nF= -(-411×10^3/96500×6)

E°cell= 0.7098V

But for Al3+(aq)/Al(s) half cell, E°= -1.66V from standard table of reduction potentials.

E°cell= E°cathode- E°anode but Al3+(aq)/Al(s) half cell is the cathode

Hence

E°anode=E°cathode - E°cell

E°anode= -1.66-0.7098= -2.37V

This is the reduction potential of Mg hence the anode material was magnesium metal

5 0

3 years ago

How do water molecules behave?

bulgar [2K]

Water is polar meaning that it has prominent areas of negative and positive charge. Among water molecules there is a weak force called hydrogen bonding which causes the molecules to "stick" to one another. This sticking to itself is called cohesion, and it causes water to ball up into droplets. This same force also causes water to stick to other kinds of molecules. This is called adhesion, and it is what causes water to ride up on the side of a water container slightly.

6 0

3 years ago

When water vapor (a gas) starts to

lisabon 2012 [21]

Idek I just need free points so I can use this app for free

8 0

4 years ago

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

Draw the product formed when 2-butanol undergoes reaction with TsCl and Et3N. CH3C6H4SO2Cl.

Pie

Answer:

Explanation:

In organic chemistry, the reaction between 2-butanol with TsCl and Et3N is known as the tosylation of the alcohol hydroxyl group. Alcohol is being changed to tosylate by the use of tosyl chloride under the influence of a base. Tosylation of alcohol is an example of a nucleophilic substitution reaction. From the image attached below, we will see how the reaction between 2-butanol proceed into the product by using tosyl chloride and a base(Et3N).

4 0

4 years ago