Answer
(a) 10.6 mA
(b) Current flows in the direction of positive charge. Therefore, current flows with the sodium ions toward
the negative electrode.
Work Step by Step
For (a):
The total charge is $(n_{Cl^{-}} + n _{Na^{+}})e = (3.92 \times 10^{16} + 2.68 \times 10^{16})(1.6 \times 10^{-19}) = 0.0106 \ C$.
Then the current $i = \frac{q}{t} = \frac{0.0106}{1} = 0.0106 \ A = 10.6 \ mA$.
