Question Analysis: The question asks for the relationship between the electromotive force ($\varepsilon$) of a cell and the potentials developed at the electrodes when no current is drawn (open circuit condition).
Step 1 (Governing Definition): [0.5 Mark]
The electromotive force ($\varepsilon$) of a cell is defined as the potential difference between the two electrodes when no current is drawn from the cell. Given the potentials at the positive electrode ($V_+$) and the negative electrode ($-V_-$) relative to the electrolyte, the total potential difference across the terminals is the algebraic difference between these potentials.
Step 2 (Mathematical Working): [0.25 Mark]
The potential of the positive electrode is $V_+$ and the potential of the negative electrode is $-V_-$. The potential difference ($\varepsilon$) is calculated as:
$$\varepsilon = V_{\text{positive}} - V_{\text{negative}}$$Substituting the given values:
$$\varepsilon = V_+ - (-V_-)$$ $$\varepsilon = V_+ + V_-$$Since both $V_+$ and $V_-$ are magnitudes of potential developed due to chemical action, their sum must be positive for the cell to function as a source of energy.
Step 3 (Final Concluded Answer): [0.25 Mark]
Therefore, the correct relationship is:
$$\varepsilon = V_+ + V_- > 0$$Correct Option: (A)
💡 EL10 Concept Insight: Think of the cell like a pump: the positive side pushes electricity out and the negative side pulls it in. The total "pushing power" (EMF) is simply the sum of how hard the positive side pushes and how hard the negative side pulls.