高考化學(xué)一輪優(yōu)化探究練習(xí)：章末排查練（六） Word版含解析

這是一份高考化學(xué)一輪優(yōu)化探究練習(xí)：章末排查練（六） Word版含解析，共6頁。
章末排查練(六)一、原電池電極反應(yīng)式或總反應(yīng)式的書寫1．鐵鎳電池(負(fù)極－Fe，正極－NiO2，電解液－KOH溶液)已知：Fe＋NiO2＋2H2OFe(OH)2＋Ni(OH)2，則：負(fù)極：________________________________________________________________________；正極：________________________________________________________________________；陰極：________________________________________________________________________；陽極：________________________________________________________________________。答案：Fe－2e－＋2OH－===Fe(OH)2　NiO2＋2H2O＋2e－===Ni(OH)2＋2OH－　Fe(OH)2＋2e－===Fe＋2OH－　Ni(OH)2－2e－＋2OH－===NiO2＋2H2O2．LiFePO4電池(正極－LiFePO4，負(fù)極－Li，含Li＋導(dǎo)電固體為電解質(zhì))已知：FePO4＋LiLiFePO4，則負(fù)極：________________________________________________________________________；正極：________________________________________________________________________；陰極：________________________________________________________________________；陽極：________________________________________________________________________。答案：Li－e－===Li＋　FePO4＋Li＋＋e－===LiFePO4　Li＋＋e－===Li　LiFePO4－e－===FePO4＋Li＋3．高鐵電池(負(fù)極－Zn，正極－石墨，電解質(zhì)為浸濕的固態(tài)堿性物質(zhì))已知：3Zn＋2K2FeO4＋8H2O3Zn(OH)2＋2Fe(OH)3＋4KOH，則：負(fù)極：________________________________________________________________________；正極：________________________________________________________________________；陰極：________________________________________________________________________；陽極：________________________________________________________________________。答案：3Zn－6e－＋6OH－===3Zn(OH)2　2FeO＋6e－＋8H2O===2Fe(OH)3＋10OH－　3Zn(OH)2＋6e－===3Zn＋6OH－　2Fe(OH)3－6e－＋10OH－===2FeO＋8H2O4．氫氧燃料電池(1)電解質(zhì)是KOH溶液(堿性電解質(zhì))負(fù)極：________________________________________________________________________；正極：________________________________________________________________________；總反應(yīng)方程式：________________________________________________________________________。(2)電解質(zhì)是H2SO4溶液(酸性電解質(zhì))負(fù)極：________________________________________________________________________；正極：________________________________________________________________________；總反應(yīng)方程式：________________________________________________________________________。(3)電解質(zhì)是NaCl溶液(中性電解質(zhì))負(fù)極：________________________________________________________________________；正極：________________________________________________________________________；總反應(yīng)方程式：________________________________________________________________________。答案：(1)2H2－4e－＋4OH－===4H2O　O2＋2H2O＋4e－===4OH－　2H2＋O2===2H2O　(2)2H2－4e－===4H＋　O2＋4H＋＋4e－===2H2O　2H2＋O2===2H2O　(3)2H2－4e－===4H＋　O2＋2H2O＋4e－===4OH－　2H2＋O2===2H2O5．甲烷燃料電池(鉑為兩極，正極通入O2和CO2，負(fù)極通入甲烷，電解液有三種)(1)電解質(zhì)是熔融碳酸鹽(K2CO3或Na2CO3)正極：________________________________________________________________________；負(fù)極：________________________________________________________________________；總反應(yīng)方程式：________________________________________________________________________。(2)酸性電解質(zhì)(電解液為H2SO4溶液)正極：________________________________________________________________________；負(fù)極：________________________________________________________________________；總反應(yīng)方程式：________________________________________________________________________。(3)堿性電解質(zhì)(電解液為KOH溶液)正極：________________________________________________________________________；負(fù)極：________________________________________________________________________；總反應(yīng)方程式：________________________________________________________________________。答案：(1)2O2＋8e－＋4CO2===4CO　CH4－8e－＋4CO===5CO2＋2H2O　CH4＋2O2===CO2＋2H2O　(2)2O2＋8e－＋8H＋===4H2O　CH4－8e－＋2H2O===CO2＋8H＋　CH4＋2O2===CO2＋2H2O　(3)2O2＋8e－＋4H2O===8OH－　CH4－8e－＋10OH－===CO＋7H2O　CH4＋2O2＋2OH－===CO＋3H2O6．甲醇燃料電池 (1)堿性電解質(zhì)(鉑為兩極，電解液為KOH溶液)正極：________________________________________________________________________；負(fù)極：________________________________________________________________________；總反應(yīng)方程式：________________________________________________________________________。(2)酸性電解質(zhì)(鉑為兩極，電解液為H2SO4溶液)正極：________________________________________________________________________；負(fù)極：________________________________________________________________________；總反應(yīng)方程式：________________________________________________________________________。答案：(1)3O2＋12e－＋6H2O===12OH－　2CH3OH－12e－＋16OH－===2CO＋12H2O　2CH3OH＋3O2＋4KOH===2K2CO3＋6H2O　(2)3O2＋12e－＋12H＋===6H2O　2CH3OH－12e－＋2H2O===12H＋＋2CO2　2CH3OH＋3O2===2CO2＋4H2O7．CO燃料電池(總反應(yīng)方程式均為2CO＋O2===2CO2)(1)熔融鹽(鉑為兩極，Li2CO3和Na2CO3的熔融鹽做電解質(zhì)，CO為負(fù)極燃?xì)?，空氣與CO2的混合氣為正極助燃?xì)?/span>)正極：________________________________________________________________________；負(fù)極：________________________________________________________________________。(2)酸性電解質(zhì)(鉑為兩極，電解液為H2SO4溶液)正極：________________________________________________________________________；負(fù)極：________________________________________________________________________。答案：(1)O2＋4e－＋2CO2===2CO　2CO＋2CO－4e－===4CO2　(2)O2＋4e－＋4H＋===2H2O　2CO－4e－＋2H2O===2CO2＋4H＋二、電解池電極反應(yīng)式和總反應(yīng)式的書寫1．用惰性電極電解下列溶液(1)NaCl溶液陰極：________________________________________________________________________；陽極：________________________________________________________________________；總反應(yīng)式：________________________________________________________________________。(2)CuSO4溶液陰極：________________________________________________________________________；陽極：________________________________________________________________________；總反應(yīng)式：________________________________________________________________________。答案：(1)2H＋＋2e－===H2↑　2Cl－－2e－===Cl2↑　2NaCl＋2H2O2NaOH＋H2↑＋Cl2↑　(2)2Cu2＋＋4e－===2Cu　4OH－－4e－===2H2O＋O2↑　2CuSO4＋2H2O2Cu＋2H2SO4＋O2↑2．用惰性電極電解下列熔融態(tài)物質(zhì)(1)MgCl2陽極：________________________________________________________________________；陰極：________________________________________________________________________；總反應(yīng)式：________________________________________________________________________。(2)Al2O3陽極：________________________________________________________________________；陰極：________________________________________________________________________；總反應(yīng)式：________________________________________________________________________。答案：(1)2Cl－－2e－===Cl2↑　Mg2＋＋2e－===Mg　MgCl2(熔融)Mg＋Cl2↑　(2)6O2－－12e－===3O2↑　4Al3＋＋12e－===4Al　2Al2O3(熔融)4Al＋3O2↑3．用銅做電極電解下列溶液(1)H2SO4溶液陰極：________________________________________________________________________；陽極：________________________________________________________________________；總反應(yīng)式：________________________________________________________________________。(2)NaOH溶液陰極：________________________________________________________________________；陽極：________________________________________________________________________；總反應(yīng)式：________________________________________________________________________。答案：(1)2H＋＋2e－===H2↑　Cu－2e－===Cu2＋　Cu＋H2SO4CuSO4＋H2↑　(2)2H2O＋2e－===H2↑＋2OH－　Cu－2e－＋2OH－===Cu(OH)2↓　Cu＋2H2OCu(OH)2↓＋H2↑4．用Al做電極電解下列溶液(1)H2SO4溶液陰極：________________________________________________________________________；陽極：________________________________________________________________________；總反應(yīng)式：________________________________________________________________________。(2)NaOH溶液陰極：________________________________________________________________________；陽極：________________________________________________________________________；總反應(yīng)式：________________________________________________________________________。答案：(1)6H＋＋6e－===3H2↑　2Al－6e－===2Al3＋　2Al＋3H2SO4Al2(SO4)3＋3H2↑　(2)6H2O＋6e－===3H2↑＋6OH－　2Al－6e－＋8OH－===2AlO＋4H2O　2Al＋2H2O＋2NaOH2NaAlO2＋3H2↑