Question

Explain Reduction on the basis of position of metals on Reactivity series ​

Answer 1

Answer:

ᴡʜᴇɴ ᴛʜᴇ ᴍᴇᴛᴀʟꜱ ᴏꜰ ᴍᴇᴅɪᴜᴍ ʀᴇᴀᴄᴛɪᴠɪᴛʏ ꜱᴇʀɪᴇꜱ ᴄᴏᴍᴇ ɪɴ ᴛʜᴇ ꜰᴏʀᴍ ᴏꜰ ᴏxɪᴅᴇ ᴛʜᴇʏ ᴀʀᴇ ᴅᴏɪɴɢ ᴀᴜᴛᴏ ʀᴇᴅᴜᴄᴛɪᴏɴ ᴘʀᴏᴄᴇꜱꜱ ᴛᴏ ɢᴇᴛ ᴘᴜʀᴇ ᴍᴇᴛᴀʟ. ᴛʜᴇ ɢᴇᴛ ᴛʜᴇ ᴍᴇᴛᴀʟ ᴏxɪᴅᴇ ꜰʀᴏᴍ ʀᴏᴀꜱᴛɪɴɢ ᴀɴᴅ ᴄᴀʟᴄɪɴᴀᴛɪᴏɴ. ... ɪɴ ᴛʜᴇ ᴀᴜᴛᴏ ʀᴇᴅᴜᴄᴛɪᴏɴ ᴘʀᴏᴄᴇꜱꜱ ᴡᴇ ɢᴇɴᴇʀᴀʟʟʏ ᴜꜱᴇ ᴄᴀʀʙᴏɴ. 8. ᴡʜᴇɴ ʀᴇᴀᴄᴛɪᴏɴ ʙᴇᴛᴡᴇᴇɴ ʜᴄꜰ ᴀɴᴅ ɴᴀ ᴛᴏᴏᴋ ᴘʟᴀᴄᴇ.

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Answer 2

In the reactivity series, as we move from bottom to top, the reactivity of metals increases. Metals present at the top of the series can lose electrons more readily to form positive ions and corrode or tarnish more readily. They require more energy to be separated from their ores, and become stronger reducing agents, while metals present at the bottom of the series are good oxidizing agent.

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