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help with spontaneous cell reactions
Hey guys am having trouble finding a way how to answer this question. I have tried working it out by writing half equations but i just dont seem to get the right answer. Could anyone possibly show me how to work it out.
he cell represented below was set up under standard conditions.
Pt|H2SO3(aq), SO4 2–(aq)||MnO4–(aq), Mn2+(aq)|Pt
write an equation for the spontaneous cell reaction.Cell
he cell represented below was set up under standard conditions.
Pt|H2SO3(aq), SO4 2–(aq)||MnO4–(aq), Mn2+(aq)|Pt
write an equation for the spontaneous cell reaction.Cell
dimz
Hey sara as this is aqa I can help. First though I need the emf values. Have they provided you with any?
The way to work it out is bascially that they give you two emf values, one for H2SO3 TO SO4 and another for MNO4- TO MN2+. Then the one with the more negative potential moves backwards. Anyway if you can provide me with the emf I will explain better,
The way to work it out is bascially that they give you two emf values, one for H2SO3 TO SO4 and another for MNO4- TO MN2+. Then the one with the more negative potential moves backwards. Anyway if you can provide me with the emf I will explain better,
Thank You, this may sound stupid but you've just helped me figured out how to work out spontaneous cell reactions.
On the LHS side of the cell what you have occuring is...
H2SO3 --> SO4(2-)
and the RHS you have...
MnO4(-) --> Mn(2+)
All you need to do now is write these conversions out as half-equations (i.e. by balancing, element, oxygens, hydrogens and charge) in the usual way...
H2SO3 + H2O --> SO4(2-) + 4H(+) + 2e(-)
and
MnO4(-) + 8H(+) + 5e(-) --> Mn(2+) + 4H2O
(notice how one is an oxidation, the other a reduction)
Then add the two together by balancing the electrons... (times first eqn by 5 and the second eqn by 2)
That leads to... (once you cancel out the H2O's and H(+)'s)
5H2SO3 + 2MnO4(-) --> 5SO4(2-) + 2Mn(2+) + 6H2O + 4H(+)
Phew!
H2SO3 --> SO4(2-)
and the RHS you have...
MnO4(-) --> Mn(2+)
All you need to do now is write these conversions out as half-equations (i.e. by balancing, element, oxygens, hydrogens and charge) in the usual way...
H2SO3 + H2O --> SO4(2-) + 4H(+) + 2e(-)
and
MnO4(-) + 8H(+) + 5e(-) --> Mn(2+) + 4H2O
(notice how one is an oxidation, the other a reduction)
Then add the two together by balancing the electrons... (times first eqn by 5 and the second eqn by 2)
That leads to... (once you cancel out the H2O's and H(+)'s)
5H2SO3 + 2MnO4(-) --> 5SO4(2-) + 2Mn(2+) + 6H2O + 4H(+)
Phew!
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