A current of two amperes is flowing through a cell of e.m.f. 5 volts and internal resistance 0.5 ohm from negative to positive electrode. If the potential of negative electrode is 10 V , the potential of positive electrode will be

Options

A. 5V
B. 14V
C. 15V
D. 16V

Answer

To solve the problem, we will use the concept of electromotive force (e.m.f.), internal resistance, and Ohm's law. Here’s the step-by-step solution:

### Step 1: Understand the given values
- Current (I) = 2 A
- E.m.f. (E) = 5 V
- Internal resistance (r) = 0.5 Ω
- Potential of negative electrode (V_neg) = 10 V

### Step 2: Calculate the voltage drop across the internal resistance
Using Ohm's law, the voltage drop (V_drop) across the internal resistance can be calculated as:
\[ V_{\text{drop}} = I \times r \]
Substituting the values:
\[ V_{\text{drop}} = 2 \, \text{A} \times 0.5 \, \Omega = 1 \, \text{V} \]

### Step 3: Relate the voltages using the e.m.f.
The potential difference between the positive and negative electrodes can be expressed as:
\[ V_{\text{pos}} = V_{\text{neg}} - V_{\text{drop}} + E \]
Where:
- \( V_{\text{pos}} \) is the potential of the positive electrode,
- \( V_{\text{neg}} \) is the potential of the negative electrode,
- \( E \) is the e.m.f.

### Step 4: Substitute the known values
Now substituting the known values into the equation:
\[ V_{\text{pos}} = 10 \, \text{V} - 1 \, \text{V} + 5 \, \text{V} \]
\[ V_{\text{pos}} = 10 - 1 + 5 \]
\[ V_{\text{pos}} = 14 \, \text{V} \]

### Step 5: Conclusion
Thus, the potential of the positive electrode is:
\[ V_{\text{pos}} = 14 \, \text{V} \]

### Final Answer
The potential of the positive electrode will be **14 V**.

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