InterviewSolution
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InterviewSolution
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A current of 9.65 A is passed through three different electrolytes NaCl, `AgNO_(3) and CuSO_(4)` for 30 min separately. Calculate the ratio of the metals deposited at the respective electrodes. Also find out the weights of various metals deposited at the respective electrodes |
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Answer» Quantity of electricity passed through the electrolyte = current strength `xx` time of flow = `9.65 xx 30 xx 60` coulombs `m_(Na) : m_(Ag) : m_(Cu) = E_(Na) : E_(Ag) : E_(Cu)` `E_(Na) = (23)/(1) = 23 " " (Na^(+) + e^(-) rarr Na)` `E_(Ag) = (108)/(1) = 108 (Ag^(+) + e^(-) rarr Ag)` `E_(Cu) = (63.5)/(2) 31.75 " " (Cu^(+2) + 2e^(-) rarr Cu)` `m_(Na) : m_(Ag) : m_(Cu) = 23 : 108 : 31.75` Weight of sodium deposited `= (23 xx 9.65 xx 30 xx 60)/(96500) = 4.14g` Weight of silver deposite `= (108 xx 9.65 xx 30 xx 60)/(96500) = 19.44 g` Weight of copper deposited `= (31.75 xx 9.65 xx 30 xx 60)/(96500) = 5.715g` |
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