Question

If $\text{E}^\circ \left( \text{Fe}_{(\text{aq})}^{+2} \mid \text{Fe}_{(\text{s})} \right) = -0.44 \text{ V}$ and $\text{E}^\circ \left( \text{Sn}_{(\text{aq})}^{+2} \mid \text{Sn}_{(\text{s})} \right) = -0.14 \text{ V}$
What is standard emf of cell containing the two electrodes?

• A. $+0.30 \text{ V}$
• B. $-0.30 \text{ V}$
• C. $+0.58 \text{ V}$
• D. $-0.58 \text{ V}$

Hint:

For a spontaneous cell, $\text{E}^\circ_{\text{cell}}$ must be positive. Subtract the smaller (more negative) value from the larger one.

Answer 1

The Correct Option is A

Step 1: Formula
$\text{E}^\circ_{\text{cell}} = \text{E}^\circ_{\text{cathode}} - \text{E}^\circ_{\text{anode}}$
Step 2: Identification
The cathode is the electrode with the higher reduction potential. - $\text{E}^\circ (\text{Sn}^{2+}/\text{Sn}) = -0.14 \text{ V}$ (Higher value, Cathode) - $\text{E}^\circ (\text{Fe}^{2+}/\text{Fe}) = -0.44 \text{ V}$ (Lower value, Anode)
Step 3: Calculation
$\text{E}^\circ_{\text{cell}} = (-0.14) - (-0.44)$
$\text{E}^\circ_{\text{cell}} = -0.14 + 0.44 = +0.30 \text{ V}$
Final Answer: (A)