Chem question

[Zerconia2921]

Quick question on redox I am getting confused by the coefficents want to know if i m doing this correctly.
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2Al + 3H2SO4 --> Al2(SO4)3 + 3H2

3H2SO4
Oxygen charge: -24
Sulfur charge : +18
Hydrogen charge : +6

Al2(SO4)3
Oxygen charge : -24
Sulfur charge: +18
Aluminum charge: +6

2Al
Aluminum charge : 0

So 2Al is the reducing agent and Al got oxidized?

P.S. Lakers RULE! And MJ is GOD

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[dentalplan]

Quick question on redox I am getting confused by the coefficents want to know if i m doing this correctly.
Also Seen On TopScore

2Al + 3H2SO4 --> Al2(SO4)3 + 3H2

3H2SO4
Oxygen charge: -24
Sulfur charge : +18
Hydrogen charge : +6

Al2(SO4)3
Oxygen charge : -24
Sulfur charge: +18
Aluminum charge: +6

2Al
Aluminum charge : 0

So 2Al is the reducing agent and Al got oxidized?

P.S. Lakers RULE! And MJ is GOD

what you're doing is correct. but dont forget, the positive values you listed (for example -24, +18, +6 for al2(so4)3) are the aggregate charges of those respectives elements. The numbers are NOT oxidations states. make sure u know the difference. For al2(so4)3 for instance, O has an oxidation number of -2, S has an ox # of 6, and Al has an ox # of 3. But since there are varying quantities of the various elements, teh ox #'s have to be multiplied by the respective multiplier values, like coefficients or subscripts.

plus u are right about the other thing, Al is the reducing species, and Al got oxidized. it got oxidized b/c it lost electrons, hence the increase in positive value. when electrons are lost, the marginal charge increases, and results in a more positively charged species. the reverse of this is called reduction.....the oxidation # is reduced, rather than increased. This results from an overall decrease in charge of the species.

ps. mj is god, i concur

 

[Zerconia2921]

what you're doing is correct. but dont forget, the positive values you listed (for example -24, +18, +6 for al2(so4)3) are the aggregate charges of those respectives elements. The numbers are NOT oxidations states. make sure u know the difference. For al2(so4)3 for instance, O has an oxidation number of -2, S has an ox # of 6, and Al has an ox # of 3. But since there are varying quantities of the various elements, teh ox #'s have to be multiplied by the respective multiplier values, like coefficients or subscripts.

plus u are right about the other thing, Al is the reducing species, and Al got oxidized. it got oxidized b/c it lost electrons, hence the increase in positive value. when electrons are lost, the marginal charge increases, and results in a more positively charged species. the reverse of this is called reduction.....the oxidation # is reduced, rather than increased. This results from an overall decrease in charge of the species.

ps. mj is god, i concur

Thanks dentalplan your my new friend.

 

[doc toothache]

Quick question on redox I am getting confused by the coefficents want to know if i m doing this correctly.
Also Seen On TopScore

2Al + 3H2SO4 --> Al2(SO4)3 + 3H2

It is not clear why you are counting charges as you did. In a redox reaction such as this only Al and hydrogen are important. Al goes from a valence of 0 to +3 for a total loss of 2(-3) electrons, while hydrogen goes from +1 to 0 for a total gain of 6(+1) electrons.

 

[osimsDDS]

wait can you explain doc toothace, how are you getting those number please thanks...

 

[nt4reall]

wait can you explain doc toothace, how are you getting those number please thanks...

See the attachment .Hope it help.

Never mind , the attachment file not clear .
Al has charge 0 when it stands alone . Then Al has charge +3 , that means Al lost 3 e- . However , there is 2 Al in the right side ( Al2(SO4)3). As a result :
2Al (0) -6e- --> 2Al (+3)
H2SO4 has 2 H and each H has +1 charge . When it moves to the right side , H2 has 0 charge. That means there is a gain of electron .
H (+1) +1e- --> H(0)
However , there is 2 H in the left side ( H2SO4) and 2H in the right side ( H2)
2H(+) + 2e- --> H2 (0)

1 2Al (0) -6e- --> 2Al (+3)
3 2H(+) + 2e- --> H2 (0)
2Al + 3H2SO4 --> Al2(SO4)3 + 3H2

 

[doc toothache]

wait can you explain doc toothace, how are you getting those number please thanks...

See partial explanation from nt4reall

See the attachment .Hope it help.

Al has charge 0 when it stands alone . Then Al has charge +3 , that means Al lost 3 e- . However , there is 2 Al in the right side ( Al2(SO4)3). As a result :
2Al (0) -6e- --> 2Al (+3)
H2SO4 has 2 H and each H has +1 charge . When it moves to the right side , H2 has 0 charge. That means there is a gain of electron .
H (+1) +1e- --> H(0)
However , there is 2 H in the left side ( H2SO4) and 2H in the right side ( H2)
2H(+) + 2e- --> H2 (0)

1 2Al (0) -6e- --> 2Al (+3)
3 2H(+) + 2e- --> H2 (0)
2Al + 3H2SO4 --> Al2(SO4)3 + 3H2

Ok except that there are 6 hydrogens rather than 2.

 

[nt4reall]

See partial explanation from nt4reall



Ok except that there are 6 hydrogens rather than 2.

Do you see the coefficient in front of the first line and the second line .

1 2Al (0) -6e- --> 2Al (+3)
3 2H(+) + 2e- --> H2 (0)

That means for the first line , you have to time 1 in the left and in the right side , as a result you have 2 Al for left and right.
For the second line , you have to time 3 for left and right sides , then you have 6 H in the left and 6 H in the right. ( and gain 6 electron from left to right )
Hope it's not so confused...
2Al + 3H2SO4 --> Al2(SO4)3 + 3H2

 

[doc toothache]

Do you see the coefficient in front of the first line and the second line .

1 2Al (0) -6e- --> 2Al (+3)
3 2H(+) + 2e- --> H2 (0)

That means for the first line , you have to time 1 in the left and in the right side , as a result you have 2 Al for left and right.
For the second line , you have to time 3 for left and right sides , then you have 6 H in the left and 6 H in the right. ( and gain 6 electron from left to right )
Hope it's not so confused...
2Al + 3H2SO4 --> Al2(SO4)3 + 3H2

3 2H(+) + 2e- --> H2 (0) should read
3 2H(+) + 6e---->3H2 (0)