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有人设计将两根铂丝做电极插入到KOH溶液中,然后在两极上分别通过甲烷和氧气而构成燃料电池。该电池中反应的化学方程式为:CH 4+2O 2+2KOH====K 2CO 3+3H 2O,则关于此燃料电池的下列说法中错误的是(1个电子所带电量为1.6×10 - 19C) A.通过甲烷的电极为电池的负极,通过氧气的电极为正极 B.在标准状况下,每消耗5.6 L O 2,可向外提供2.4×10 4 C的电量 C.通过甲烷电极的电极反应为:CH 4+10OH --8e -==CO 3 2-+7H 2O D.放电一段时间后,溶液的pH升高
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