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原电池原理

> 纽扣电池的两极材料分别为锌和氧化银，电解质溶液为KOH溶液。放电时两个电极反应分别为：Zn+2OH -=Zn(OH) 2 +2e -; Ag 2O +H 2O +2e - ="2Ag" +2OH -。下列说法错误的是 A．溶液中OH-向正极移动 B．锌发生氧化反应 C．氧化银是正极 D．在电池放电过程中，电解质溶液的碱性增强

问题：

发布时间：2024-09-07

纽扣电池的两极材料分别为锌和氧化银，电解质溶液为KOH溶液。放电时两个电极反应分别为：Zn+2OH ^-=Zn(OH) ₂ +2e ^-; Ag ₂O +H ₂O +2e ^- ="2Ag" +2OH ^-。下列说法错误的是

A．溶液中OH^-向正极移动 B．锌发生氧化反应

C．氧化银是正极 D．在电池放电过程中，电解质溶液的碱性增强

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