Question

The electromotive force of the following cell is Zn$|$Zn$^{2+}$ (1M) $||$ Fe$^{2+}$ (1M) $|$Fe E$^0$ Zn$^{2+}$|Zn = -0.76 V, E$^0$ Fe$^{2+}$|Fe = -0.44 V

• A. 1.2 V
• B. 0.32 V
• C. -1.2 V
• D. -0.32 V

Hint:

The EMF is positive if the cathode has a higher potential than the anode, which corresponds to a spontaneous reaction.

Answer 1

The Correct Option is B

Step 1: Understand the formula for electromotive force.
The electromotive force (EMF) of a galvanic cell is calculated using the standard electrode potentials of the two half-reactions. The formula is: \[ \text{EMF} = E^0_{\text{cathode}} - E^0_{\text{anode}} \]
Step 2: Identify the cathode and anode.
- Zinc (Zn) is the anode because it has a more negative standard electrode potential and will undergo oxidation (Zn $\rightarrow$ Zn$^{2+}$ + 2e$^-$). - Iron (Fe) is the cathode because it has a less negative standard electrode potential and will undergo reduction (Fe$^{2+}$ + 2e$^-$ $\rightarrow$ Fe).
Step 3: Apply the electrode potentials.
- Standard electrode potential for zinc (Zn$^{2+}$|Zn) is -0.76 V. - Standard electrode potential for iron (Fe$^{2+}$|Fe) is -0.44 V. So, the EMF is: \[ \text{EMF} = (-0.44) - (-0.76) = 0.32 \text{ V} \]
Final Answer: \[ \boxed{0.32 \text{ V}} \]