Question

Which from following formulae is used to obtain value of $\text{E}^\circ_{\text{cell}}$ for a reaction taking place in Dry cell?

• A. $\frac{-\Delta\text{G}^\circ}{\text{F}}$
• B. $\frac{-\Delta\text{G}^\circ}{2F}$
• C. $\frac{-\Delta\text{G}^\circ}{3 \text{F}}$
• D. $\frac{-\Delta\text{G}^\circ}{4F}$

Hint:

In these formulas, "n" is the key. Since Zinc is divalent ($Zn^{2+}$), 2 electrons are involved, making $n=2$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The standard Gibbs free energy change ($\Delta G^\circ$) of an electrochemical cell reaction is fundamentally related to the standard cell potential ($E^\circ_{\text{cell}}$) by the core thermodynamic equation $\Delta G^\circ = -nFE^\circ_{\text{cell}}$, where $n$ represents the number of moles of electrons transferred in the balanced cell reaction, and $F$ is the Faraday constant.
Step 2: Key Formula or Approach:
The rearranged formula to solve for the cell potential is: \[ E^\circ_{\text{cell}} = \frac{-\Delta G^\circ}{nF} \] Approach: Determine the value of '$n$' for the overall chemical reaction occurring specifically in a Leclanché cell (dry cell).
Step 3: Detailed Explanation:
To find the correct formula, we must identify the total electron transfer in a dry cell. The half-reactions operating in a standard dry cell are: At Anode (Oxidation of zinc container): \[ \text{Zn}(s) \longrightarrow \text{Zn}^{2+}(aq) + 2\text{e}^- \] At Cathode (Reduction of manganese dioxide): \[ 2\text{MnO}_2(s) + 2\text{NH}_4^+(aq) + 2\text{e}^- \longrightarrow \text{Mn}_2\text{O}_3(s) + 2\text{NH}_3(g) + \text{H}_2\text{O}(l) \] The balanced overall cell reaction involves the net transfer of exactly 2 electrons. Therefore, $n = 2$. Substitute $n = 2$ into the general thermodynamic relationship: \[ E^\circ_{\text{cell}} = \frac{-\Delta G^\circ}{nF} \] \[ E^\circ_{\text{cell}} = \frac{-\Delta G^\circ}{2F} \] Step 4: Final Answer:
The correct formula is $\frac{-\Delta\text{G}^\circ}{2F}$.