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 need to find the potential of the positive electrode given the current flowing through the cell, the e.m.f., the internal resistance, and the potential of the negative electrode.

### Step-by-step Solution:

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

2. **Calculate the voltage drop across the internal resistance:**
The voltage drop (V_drop) across the internal resistance can be calculated using Ohm's Law:
\[ V_{\text{drop}} = I \times r \]
Substituting the values:
\[ V_{\text{drop}} = 2 \, \text{A} \times 0.5 \, \Omega = 1 \, \text{V} \]

3. **Determine the potential of the positive electrode (V_pos):**
The potential of the positive electrode can be found using the formula:
\[ V_{\text{pos}} = V_{\text{neg}} + E - V_{\text{drop}} \]
Substituting the known values:
\[ V_{\text{pos}} = 10 \, \text{V} + 5 \, \text{V} - 1 \, \text{V} \]
\[ V_{\text{pos}} = 14 \, \text{V} \]

4. **Final Answer:**
The potential of the positive electrode is **14 V**.