A) 1.056 mA, positive to negative electrode
B) 10.56 mA, positive to negative electrode
C) 1.056 mA, negative to positive electrode
D) 10.56 mA, negative to positive electrode
Correct Answer: B
Solution :
Current\[=\frac{q}{t}=\frac{[{{n}_{N{{a}^{+}}}}+{{n}_{C{{l}^{-}}}}]e}{t}\] \[=\frac{(2.68\times {{10}^{16}}+3.92\times {{10}^{16}})(1.6\times {{10}^{-19}})}{1.00}\] =10.56 mA Direction of current is from positive electrode to negative electrode as Na+ are positively charged while Cl- are negatively charged.You need to login to perform this action.
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