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Samenvatting H18 (scheikunde) - Accu's en brandstoffen Chemie Overal Vwo 6 $3.20
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Samenvatting H18 (scheikunde) - Accu's en brandstoffen Chemie Overal Vwo 6
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  • Course
  • Scheikunde
  • Level
  • VWO / Gymnasium
  • Book
  • Chemie overal 6v

Samenvatting van hoofdstuk 18 uit het boek Chemie Overal voor VWO 6 over accu's en brandstoffen. Begrippen staan dikgedrukt. Ondersteund met plaatjes. Duidelijk en uitgebreid. Mocht er iets niets duidelijk zijn mag je me altijd een berichtje sturen.

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  • No
  • H18
  • March 7, 2022
  • 3
  • 2021/2022
  • Summary

Subjects

  • accu
  • brandstoffen
  • scheikunde
  • vwo

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  • Secondary school
  • VWO / Gymnasium
  • Scheikunde
  • 6
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H18 – ACCU’S EN
BRANDSTOFCELLEN
18.1 – ELEKTROMAGNETISCHE CEL


STROOM UIT STOFFEN
 Elektrochemische cel zet via redoxreacties chemische energie om in elektrische energie
 Oxidator/reductor in losse vaten (halfcellen) verbonden met stroomdraden en zoutbrug
1) Reductor staat elektronen af > negatieve elektrode > elektronen door stroomdraad naar oxidator
2) Oxidator neemt op > positieve elektrode
 Zoutbrug (sluit stroomkring) gevuld met elektrolyt (zout): ionen naar reductor(-)/oxidator(+)
 Alternatief: Membraan die selectief stoffen doorlaat en wel contact verhinderd


BRONSPANNING
 Bronspanning = verschil in standaardelektronenpotentiaal (V0) = V0 Oxidator – V0 Reductor

18.2 – BATTERIJ EN BRANDSTOF


BATTERIJ
 Batterij – MnO2 (oxidator) en Zn (reductor) gescheiden door NH4Cl (zoutbrug)
1) MnO2 (s) + H2O (l) + e-  MnO(OH) (s) + OH-
2) Zn (s)  Zn2+ + 2 e-


OPLAADBARE BATTERIJ
 Bij wegwerpbatterijen lossen ionen op in elektrolyt > niet oplaadbaar
 Bij oplaadbare batterij blijven ionen bij elektrode > opladen
 Aansluiten aan sterkere spanningsbron > reactie verloopt andersom = elektrolyse
 Resterende stoffen in reactievergelijking zijn vast of in zoutbrug > ionen blijven bij
elektrode


BRANDSTOFCEL
 Brandstofcel is kleiner/lichter dan accu en bevat meer energie
 In 1 halfcel wordt continu O2 (oxidator) geleid, in de ander brandstof (reductor)
 Membraan laat H+ door zodat O2 water kan vormen

1) O2 (g) + 4 H+ + 4 e-  2 H2O (l) (3x)
2) CH3OH (l) + H2O (l)  CO2 (g) + 6 H+ + 6 e- (2x)
 H+ en H2O wegstrepen

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