Question

Calculate the amount of copper metal deposited at the cathode when I.5 ampere current was passed for one hour in a solution of copper sulphate.​

Answer 1

$\huge \mathsf \fcolorbox {magenta}{lavender}{Answer}$

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➩ ᴛʜɪs ɪs ʙᴀsᴇᴅ ᴏɴ ғᴀʀᴀᴅᴀʏ's ʟᴀᴡ ᴏғ ᴇʟᴇᴄᴛʀᴏʟʏsɪs

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$\sf \pink {\fbox {Given:-}}$

$➩{Current (I) = 1.5amperes}$

$➩{Time (t) = 1hour= 60 min}$

$➩{Time(t)=60×60=>3600s}$

$➩{Total\:charge(Q) = 1.5×3600}$

$➩{Total\:charge(Q) =5400columbs}$

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$\sf \orange {\fbox {we\: know,}}$

$➩{n=\frac{Q}{zF}}$

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$\sf \purple {\fbox {where,}}$

➩ n= ɴᴜᴍʙᴇʀ ᴏғ ᴍᴏʟᴇs ᴏғ ᴄᴏᴘᴘᴇʀ ᴘʟᴀᴛᴇᴅ ᴏᴜᴛ

$➩{F\: is\: Faraday's\: constant (F)}$

$➩{The\: value\: of\: F=96,485/mol}$

$\sf \red {\fbox {And}}$

➩ ᴢ= ɴᴜᴍʙᴇʀ ᴏғ ᴇʟᴇᴄᴛʀᴏɴs ɪɴ ʜᴀʟғ-ᴄᴇʟʟ ʀᴇᴀᴄᴛɪᴏɴ=>2

$➩{Total\: charge\: Q=5400c}$

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$\sf \green {\fbox {On\: solving}}$

$➩{n=\frac{5400}{(2×96,485)}}$

$➩{n=\frac{5400}{192970}}$

$➩{n=0.027mol}$

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$➩{n=\frac{Weight\: of\: cu}{molecular\: wt\: of\: cu}}$

$➩{0.027=\frac{Weight\: of\: cu}{63.5}}$

$➩{Weight\: of\: cu=63.5×0.027}$

$➩{Weight\: of\: cu=1.714grams}$

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$\sf \blue {\fbox {Therefore}}$

➩The amount of copper metal deposited at cathode=1.714grams

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$\mathsf \fcolorbox {lavender}{lavender}{Hope\: it\: works}$