In an electrolyte 3-2 x 1018 bivalent positive ions drift to the right per second while 3-6 x 1018 monovalent negative ions drift to the left per second- Then the current will be

Current(I) = net charge(Q) /timeNet charge Q = (2* 3.2* 1018 3.6* 1018)e nbsp;= 2.8 *1018*1.6*10 19Q = 0.448 Ct =1sSo, I = (0.448/1) = 0.448 A. ...

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Standard XII

Chemistry

Electrical Properties of Colloids

In an electro...

Question

In an electrolyte 3.2 x 10¹⁸ bivalent positive ions drift to the right per second while 3.6 x 10¹⁸ monovalent negative ions drift to the left per second. Then the current will be

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Solution

Current(I) = net charge(Q) /time
Net charge Q = (2* 3.2* 10¹⁸ - 3.6* 10¹⁸)e = 2.8 10¹⁸1.6*10^-19
Q = 0.448 C
t =1s
So, I = (0.448/1) = 0.448 A.

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In an electrolyte 3.2 x 1018 bivalent positive ions drift to the right per second while 3.6 x 1018 monovalent negative ions drift to the left per second. Then the current will be

Current(I) = net charge(Q) /timeNet charge Q = (2* 3.2* 1018 - 3.6* 1018)e = 2.8 *1018*1.6*10-19Q = 0.448 Ct =1sSo, I = (0.448/1) = 0.448 A.

In an electrolyte 3.2 x 10¹⁸ bivalent positive ions drift to the right per second while 3.6 x 10¹⁸ monovalent negative ions drift to the left per second. Then the current will be

Current(I) = net charge(Q) /time
Net charge Q = (2* 3.2* 10¹⁸ - 3.6* 10¹⁸)e = 2.8 10¹⁸1.6*10^-19
Q = 0.448 C
t =1s
So, I = (0.448/1) = 0.448 A.