In an electrolyte,3.2 times 10^{18} bivalent | vedclass

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(B) The current due to positive ions moving to the right is $i_{(+)} = \frac{n_{(+)} q_{(+)}}{t} = (3.2 	imes 10^{18}) 	imes (2e) / 1 \, s = 6.4 	i

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$1.6 \, A$ to the right

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(B) The current due to positive ions moving to the right is $i_{(+)} = \frac{n_{(+)} q_{(+)}}{t} = (3.2 	imes 10^{18}) 	imes (2e) / 1 \, s = 6.4 	imes 10^{18} 	imes 1.6 	imes 10^{-19} = 1.024 \, A$ (to the right).The current due to negative ions moving to the left is equivalent to a current of positive charge moving to the right. Thus,$i_{(-)} = \frac{n_{(-)} q_{(-)}}{t} = (3.6 	imes 10^{18}) 	imes (e) / 1 \, s = 3.6 	imes 10^{18} 	imes 1.6 	imes 10^{-19} = 0.576 \, A$ (to the right).The net current is $i_{net} = i_{(+)} + i_{(-)} = 1.024 \, A + 0.576 \, A = 1.6 \, A$ (to the right).

In an electrolyte,$3.2 \times 10^{18}$ bivalent positive ions drift to the right per second,while $3.6 \times 10^{18}$ monovalent negative ions drift to the left per second. Then the current is:

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