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: Understanding the Concept:
The standard emf (\(\text{E}^\circ_{\text{cell}}\)) of a galvanic cell is the difference between the standard reduction potentials of the cathode and the anode.
A spontaneous reaction has a positive cell potential.
Step 2: Key Formula or Approach:
\[ \text{E}^\circ_{\text{cell}} = \text{E}^\circ_{\text{cathode}} - \text{E}^\circ_{\text{anode}} \] where the electrode with the higher (more positive or less negative) reduction potential acts as the cathode.
Step 3: Detailed Explanation:
Given standard reduction potentials:
\(\text{E}^\circ(\text{Sn}^{+2}|\text{Sn}) = -0.14 \text{ V}\)
\(\text{E}^\circ(\text{Fe}^{+2}|\text{Fe}) = -0.44 \text{ V}\)
Since \(-0.14 \text{ V}>-0.44 \text{ V}\), the Tin electrode (\(\text{Sn}\)) is the cathode and the Iron electrode (\(\text{Fe}\)) is the anode.
\[ \text{E}^\circ_{\text{cell}} = (-0.14 \text{ V}) - (-0.44 \text{ V}) \] \[ \text{E}^\circ_{\text{cell}} = -0.14 + 0.44 = +0.30 \text{ V} \] Step 4: Final Answer:
The standard emf of the cell is \(+0.30 \text{ V}\).