想問2012practice paper1BQ8(direct methanol fuel cell)嗰兩條half equation係要自己deduce定有得背 ?
要deduce嘅話有無方法
有
fuel=有oxygen
咁黎手寫番條oxygen既式出黎先喇
O2+2H2O+4e- -->4OH-
我冇題目係手所以估係acidic medium
O2+2H2O+4H+ + 4e- -->4H2O
O2+4H+ + 4e- --> 2H2O
咁哩個收電子
姐係有個要放電子
姐係CH3OH係放電子
咪住先我唔知CH3OH放電子之後product出咩喎
再睇番題目,係fuel cell
咩係fuel呀?
姐係combustion
姐係overall equation我應該寫到條combustion既式
e.g CH3OH + 3/2O2 --> CO2 + 2H2O
2CH3OH+3O2-->2CO2+4H2O
我依家得CO2未有,姐係CH3OH出CO2
姐係CH3OH-->CO2
再用H+,H2O balance下
會出到 CH3OH +H2O--> CO2 +6H+6e-
好詳盡
唔該晒巴打
呢題我每年教到cell一定教埋呢題
2CH3OH + 3O2 -> 2CO2 + 4H2O
首先你要識抽返anode同cathode既reaction出黎先
根據oxidation number, CH3OH -> CO2
O.N. of C: -2 -> +4, so oxidation (anode)
O.N. of O: 0 -> -2, so reduction (cathode)
anode: CH3OH -> CO2
cathode: O2 -> H2O
用返平時普通bal redox既方法就得 (補H2O,H+,e-)
anode:CH3OH + H2O -> CO2 + 6H+ + 6e-
cathode: O2 + 4H+ + 4e- -> 2H2O
如果bal完唔sure啱唔啱 可以用O.N.驗算
因為C個O.N. total升左6, 所以anode條式會放6個e-就啱
O個O.N. total跌4(因為O2兩個O reduce), 所以cathode條式會收4個e-就啱
題目講明acidic medium所以唔洗用OH-去抵消d H+
syllabus學果7隻電池既equation全部都唔洗背
100%可以用呢個方法prove返出黎