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Chemistry: Post your doubts here!

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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
 
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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
 
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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
 
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#9) idk :p


#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? :unsure:
 
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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 ?
 
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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!
 
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