In an electrolyte `3.2 xx 10^(18)` bivalent positive ions drift to the right per second while `3.6 xx 10^(18)` monovalent negaitve ions drift to the per second. Then the current is

Options

A. 1.6 amp to the left
B. 1.6 amp to the right
C. 0.45 amp to the right
D. 0.45 amp to the left

Explanation

To find the current in the electrolyte given the drift of bivalent positive ions and monovalent negative ions, we can follow these steps:

### Step 1: Understand the charges of the ions
- Bivalent positive ions have a charge of \( +2e \) (where \( e \) is the elementary charge, approximately \( 1.6 \times 10^{-19} \) coulombs).
- Monovalent negative ions have a charge of \( -e \).

### Step 2: Identify the number of ions drifting
- The number of bivalent positive ions drifting to the right per second is \( N_+ = 3.2 \times 10^{18} \).
- The number of monovalent negative ions drifting to the left per second is \( N_- = 3.6 \times 10^{18} \).

### Step 3: Calculate the total charge contributed by the bivalent positive ions
- The total charge due to the bivalent positive ions is given by:
\[ Q_+ = N_+ \times (2e) = (3.2 \times 10^{18}) \times (2 \times 1.6 \times 10^{-19}) \]
- Simplifying this:
\[ Q_+ = 3.2 \times 10^{18} \times 3.2 \times 10^{-19} = 10.24 \times 10^{-1} = 1.024 \text{ C/s} \]

### Step 4: Calculate the total charge contributed by the monovalent negative ions
- The total charge due to the monovalent negative ions is given by:
\[ Q_- = N_- \times (-e) = (3.6 \times 10^{18}) \times (-1.6 \times 10^{-19}) \]
- Simplifying this:
\[ Q_- = -3.6 \times 10^{18} \times 1.6 \times 10^{-19} = -5.76 \times 10^{-1} = -0.576 \text{ C/s} \]

### Step 5: Calculate the net current
- The net current \( I \) is the sum of the contributions from both types of ions:
\[ I = Q_+ + Q_- = 1.024 - 0.576 = 0.448 \text{ A} \]

### Step 6: Determine the direction of the current
- Since the bivalent positive ions contribute more charge than the monovalent negative ions, the net current will be directed to the right.

### Final Answer
The net current in the electrolyte is \( 0.448 \) A towards the right.

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