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CHEM PPR 4 HELP! URGENT!!!!!

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Jun10/42...Q8b!!!

Howcum the voltage cud be 1.6V???????????? :shock:

Explain plzzzz!!!!!! and about calculating volts given equations, i also dunno the rule, wud be thankful if sonmeone cud help!!!!!!!!!!! :)
 
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darkakash said:
Its Easy.....
Higher value - lower value....


like what??? how did u do?? what values??? i tried...its not cuming 1.6V!!!
please can u solve here??thnks
 
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someone plzzzz explain me N09/42.......Q8ci and cii. i dont get how u find the ratio of the peaks?????
 
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darkakash said:
Its Easy.....
Higher value - lower value....

Reaction I: Pb + SO4---> PbSO4 + 2e–
Reaction II: PbO2 + 4H+ + SO4– + 2e ---->PbSO4 + 2H2O

In reaction 1 and 2 we neglect "SO4" as it is nor giving or recieving electrons. So now write both the equation without SO4.....
Reaction I: Pb---> Pb* + 2e–
Reaction II: PbO2 + 4H* + 2e ---->Pb* + 2H2O {* is +ve charge}

E value for reaction I = -0.13
E value foe rwaction II = +1.47

Higher value - lower value = 1.47 - (-0.13) = +1.60

Hope u Got it : :D :D
 
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darkakash wow you are great help :)

but do we apply this rule (higher value - lower value) to all Electrochemistry questions?
 

Xam

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@max its the rule man v always subtract higher - lower potential
 
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no electrochem questions came anyway! I didn't study it much. I read the chapter in the big Cambridge text book but that wasn't enough. I didn't study much this year :(
But thank GOD I did well in my paper 5 chem.
 
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