(1) 2CH 3OH(1) +3O 2(g) = 2CO 2(g) + 4H 2O(g);△H =-1453.28kJ·mol - 1
(2)负极;电极反应方程式为O 2+4e -+2H 2O=4OH -
(3) 1.07 g。
CH 3OH(1) + O 2(g) =" CO(g)" + 2H 2O(g)△H1= -443.64 kJ·mol - 1
2CO (g) + O 2(g) = 2CO 2(g) △H 2=-566.0 kJ·mol - 1
2CH 3OH(1) +3O 2(g) = 2CO 2(g) + 4H 2O(g);△H =2△H1-△H 2=-1453.28kJ·mol - 1
还原剂做负极,氧化剂做正极,O 2+4e -+2H 2O=4OH -
铜的物质的量为0.1mol , 至少消耗甲醇的质量为1.07g