Question

In an ionic solution, a current consists of Ca2+Ca2+ ions (of charge +2e+2e) and Cl−Cl− ions (of charge −e−e) traveling in opposite directions. Part A If 5.02×1018 Cl−Cl− ions go from A to B every 0.620 minmin , while 3.25×1018 Ca2+Ca2+ ions move from B to A, what is the current (in mAmA) through this solution? Part B In which direction (from A to B or from B to A) is the current in part A going?

Answer 1

Answer:

A

The current through the solution is   $I =35.61mA$

B

The current is moving from B to A

Explanation:

From the question we are told that

      The number of  $Cl^{-}$ that move from A to B is $N_{cl} = 5.02 *10^{18} ions$

      The time taken to move from A to B   $t_{cl} = 0.620$

                                                                          $=37.2 s$

Since the value of 1 charge is  $q =1.602 *10^{-19} C$

The quantity of charge Q that flow from A to B is mathematically given as

                         $Q_{cl} = 5.02 *10^{18} * 1.602*10^{-19}$

                                $=0.804C$

   The number of  $Ca^{+}$ that move from A to B is $N_{ca} = 3.25*10^{18} ions$

      Since time taken to move from A to B is equal to time taken to move from B to A   $t =t_{cl} =t_{ca}= 0.620=37.2s$

 

The quantity of charge Q that flow from B to A is mathematically given as

                         $Q_{ca} = 3.25 *10^{18} * 1.602*10^{-19}$

                                $=0.5207C$

The total quantity of charge is

                                $Q_{tot}=Q_{cl} + Q_{ca}$

                                $Q_{tot}= 0.804 + 0.5207$

                                       $= 1.325C$

The current flowing through the solution is

                         $I =\frac{Q_{tot}}{t}$

                            $I = \frac{1.325}{37.2}$

                                $= 0.03561A$

                            $I =35.61mA$

The flow is from B to A cause current  flow from the positive terminal to negative terminal