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#### RING12

##### Member
10+ Year Member
7+ Year Member

please help tomorrow is my test. my brian does not work any more. could you help me with

how many grams of NAOH(40g/mol) are there in 250 mL of 0.4M NAOH solution?

i solved befor but i can not do it again.

given the following half-cell reaction:

cl2(g)+2e______ 2cl E=01.36v
cu2+ (aq) +2e_______ cu(S) E=+0.34

answer is:-1.02 i keep getting =1.02

#### aranjuez

##### Member
10+ Year Member
7+ Year Member
RING12 said:
please help tomorrow is my test. my brian does not work any more. could you help me with

how many grams of NAOH(40g/mol) are there in 250 mL of 0.4M NAOH solution?

i solved befor but i can not do it again.

250 mL = 1/4 L
0.4M NaOH = 4 mol/ 10 L

(1/4)L * (4mol/10L) = 0.1 mol
0.1 mol * 40g/mol = 4 g

aranjuez

#### aranjuez

##### Member
10+ Year Member
7+ Year Member
2Cl --> Cl2(g)+2e E=-1.36v
Cu2+ (aq) +2e --> Cu(S) E=+0.34

Try it now.

aranjuez

#### RING12

##### Member
10+ Year Member
7+ Year Member
aranjuez said:
2Cl --> Cl2(g)+2e E=-1.36v
Cu2+ (aq) +2e --> Cu(S) E=+0.34

Try it now.

aranjuez

one question, i thought we should change the sign of oxidation reaction not the reduction reaction. the one more positive is reduction. can you tell me whcih part am i wrong?

#### Notoriousjae

##### DAT Rockstar
10+ Year Member
RING12 said:
please help tomorrow is my test. my brian does not work any more. could you help me with

how many grams of NAOH(40g/mol) are there in 250 mL of 0.4M NAOH solution?

i solved befor but i can not do it again.

given the following half-cell reaction:

cl2(g)+2e______ 2cl E=01.36v
cu2+ (aq) +2e_______ cu(S) E=+0.34

answer is:-1.02 i keep getting =1.02

1.250mL*1L/1000mL * 0.4mol/L of NaOH*40g/mol of NaOH=4

2. cl2 +2e-->2cl E=1.36
cu--->cu2+ +2e E=-0.34
E=1.02

#### Notoriousjae

##### DAT Rockstar
10+ Year Member
you are not wrong for the second answer. you have to change the sign for oxidation, and the higher E cell value is the one that is reduced so it should be positive

#### RING12

##### Member
10+ Year Member
7+ Year Member
Notoriousjae said:
you are not wrong for the second answer. you have to change the sign for oxidation, and the higher E cell value is the one that is reduced so it should be positive

then do you think the anwer sheet of ada is wrong?
b/c it's says -1.02

#### Notoriousjae

##### DAT Rockstar
10+ Year Member
RING12 said:
then do you think the anwer sheet of ada is wrong?
b/c it's says -1.02

yes the answer sheet is wrong.

#### Indie

##### Member
10+ Year Member
Notoriousjae said:
yes the answer sheet is wrong.

The answer sheet is not wrong.. If you look at the overall equation they give you cu+2 + 2cl- --> cu + cl2

in the first half cell rxn they give you 2cl is a product and cl2 is a reactant... In the final equation you need the opposite so you would reverse the rxn and have -1.36v
the second half rxn is written correctly so you just use the value +.34v
-1.36 + .34 = -1.02v

#### Notoriousjae

##### DAT Rockstar
10+ Year Member
Indie said:
The answer sheet is not wrong.. If you look at the overall equation they give you cu+2 + 2cl- --> cu + cl2

in the first half cell rxn they give you 2cl is a product and cl2 is a reactant... In the final equation you need the opposite so you would reverse the rxn and have -1.36v
the second half rxn is written correctly so you just use the value +.34v
-1.36 + .34 = -1.02v

i personally did not actually see this problem; i was only going by the half cell info the OP posted, but if they give you an overall redox reaction, then indie is correct.