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Suppose 250 mL of a 0.433 M solution of CuCl_(2) is electrolysed. How long will a current of 0.75 A have to run in order to reduce the concentration of Cu^(2+) to 0.167 M? What mass of Cu(s) will be deposited on the cathode during this time ? |
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Answer» Solution :Mole of `Cu^(2+)` reduced `= 0.433 XX 0.250 - 0.167 xx 0.250` = 0.0665. Wt. of Cu deposited `= 0.0665 xx 63.5g` Let the time to DEPOSIT this amount of Cu be t seconds `THEREFORE` no. of F = no. of EQ. `= (0.75 xx t)/(96500) = 0.0665 xx 2` `t = 1.71 xx 10^(4)` seconds. |
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