Chemistry: Post your doubts here!

[ravaneous]

Increasing the iodide concentration would increase the voltage or decrease ?

 

[A star]

Increasing the iodide concentration would increase the voltage or decrease ?

i didnt get you question? is it p4 and p5?

 

[ravaneous]

i didnt get you question?

increasing the concentration of iodide half cell connected to hydrogen would it increase the Ecell or decrease it

 

[KurayamiKimmi]

View attachment 23968
Increasing the iodide concentration would increase the voltage or decrease ?

It depends on whether iodide ion concentration is being increased or iodine concentration

If iodine concentration increased

• equilbrium shifts to right
• becomes more easy to form ions
• more positive

If iodide ion concentration is increased

• equilibrium shifts to left
• harder to form ions
• more negative

 

[Muhammad Bin Anis]

View attachment 23968
Increasing the iodide concentration would increase the voltage or decrease ?

The present value of +0.54V is given for standard conditions.
Increasing the iodine concentration will result in the equilibrium shifting forward. So the Ecell value will increase.

 

[iKhaled]

The position of an equilibrium is affected by change in concentration pressure etc .Redox equilibrium are da same.
thereforeif you increase the iodide concentration :-

• According to Le Chatelier , the equilibrium will move to oppose the change
• This means that it will try to reduce the concentration of iodide --> by increasing con of ions
• making voltage value more positive.

more positive??? why the equilibrium position will shift to the left which means the value will be less positive so it will decrease

 

[iKhaled]

View attachment 23968
Increasing the iodide concentration would increase the voltage or decrease ?

i say it will decrease the voltage because the equilibrium position will shift to the left.

 

[syed1995]

i didnt get you question? is it p4 and p5?

It's A2

 

[gary221]

oct nov 43 p4 q3bi

what year??

 

[Tkp]

what year??

haha well i progressed to the future,its oct nov 12 43

 

[syed1995]

haha well i progressed to the future,its oct nov 12 43

I am sure that question will in oct/nov 43!

 

[Tkp]

c i told you that i have time machine

I am sure that question will in oct/nov 43!

 

[KurayamiKimmi]

more positive??? why the equilibrium position will shift to the left which means the value will be less positive so it will decrease

i changed my initial answer. Is it clearer now ?

 

[iKhaled]

i changed my initial answer. Is it clearer now ?

yeah it is much more clearer now!

 

[KurayamiKimmi]

yeah it is much more clearer now!

good to know

 

[gary221]

oct nov 43 p4 q3bi

3 b i) Since H is added to the indigo (the 2 carboxyl group (ketone) r converted to -OH ) --> it is a reduction rxn!
ii) Any reagent which converts the carboxyl group to alcohol (-OH) ie a reducing agent eg NaBH4 or LiAlH4

 

[messi10]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s10_qp_41.pdf

Question #5
(a) Part (iii)
and
(b) Part (iv)

 

[Warrior66]

#9) idk


#21) organic compounds that react with sodium are either carbonal acids or alcohols ... organic ion with a charge of -3 means that there are 3 charged group or in other words three groups in the compound that could react , j reacts with NaOH to give a compound with charge -1 meaning that 2 of the charged groups will react the other wont .. thus it should have 2 acid groups and 1 alcohol group which wont react with NaOH answer=C


#26) compounds that could be oxidised by acidified dichromate are alcohols,ketones & aldehydes .. but then only aldehydes give a positive result with tollen's so the compounds should contain both alcohol aldehyde .. the only answer which works is C

#28) the graph shows a reaction that is done in 2 steps & is exothermic the only one which fits is B which is SN2 reaction

Jazak'Allahum Khairan!!
btw in q 26..you said that ketones can be oxidised by acidified dichromate? I don't think that's right, is it?

 

[iKhaled]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s10_qp_41.pdf

Question #5
(a) Part (iii)
and
(b) Part (iv)

question 5 part iii.

2H2O + 2e (EQUILIBRIUM ARROW) H2 + 2OH- E = -0.83V (left electrode)
2O2 + 2H2O + 4e (EQUILIBRIUM ARROW) 4OH- E= 0.40V (right electrode)

now E= 0.40 - (-0.83) v

now when we increase the OH concentration on the left electrode, the equilibrium position will shift to the left making the value of E more negative

now when we increase the OH concentration on the right electrode the equilibrium position will shit to the left also making the value E of the right electrode becoming less positive

did u get it ?

 

[iKhaled]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s10_qp_41.pdf

Question #5
(a) Part (iii)
and
(b) Part (iv)

for part b (iv)

at the cathode: Pb2+ + 2e (equilibrium arrow) Pb E= -0.13V
at the anode: PBO2 + 4H+ +2e (equilibrium arrow) Pb2+ + 2H2O E= 1.47V

E of the cell = 1.47 - (-0.13)

now decreasing the Pb+2 concentration on the cathode will shift the equilibrium to the left which will make E of the electrode more negative and decreasing Pb+2 concentration on the anode will shift the equilibrium position to the right which will make the value E of the electrode more positive so definitely E of the cell will be more positive

i hope you got it!