Question

At 298 K, $E^{\circ}$ value of the cell involving $2 Fe^{3+}(aq) + 2 I^-(aq) \rightarrow 2 Fe^{2+}(aq) + I_2(s)$ is X V. The X (in V) and $\log K_c$ for the reaction are respectively: ($E^{\circ}_{Fe^{3+}/Fe^{2+}} = 0.77$ V, $E^{\circ}_{I_2/I^-} = 0.54$ V)

• A. 0.23, 8.79
• B. -0.23, 9.79
• C. 0.23, 7.79
• D. -0.23, 6.79

Hint:

$\log K_c = \frac{n E^{\circ}}{0.0591}$ at 298 K!

Answer 1

The Correct Option is C

Step 1: Decide cathode and anode.
The species that gets reduced (gains electrons) is the cathode. Here $Fe^{3+}$ becomes $Fe^{2+}$, so iron is the cathode. Iodide is oxidised to iodine, so it is the anode.

Step 2: Use the cell potential formula.
The standard cell potential is \[ E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} \]

Step 3: Plug in the numbers.
$E^{\circ}_{cell} = 0.77 - 0.54 = 0.23$ V. This is positive, so the reaction is spontaneous. That gives us $X = 0.23$ V.

Step 4: Count the electrons.
Each iron gains one electron and there are two irons, so $n = 2$ electrons are transferred.

Step 5: Link potential to the equilibrium constant.
The bridge formula at 298 K is \[ \log K_c = \frac{n\,E^{\circ}_{cell}}{0.0591} \]

Step 6: Do the arithmetic.
$\log K_c = \dfrac{2 \times 0.23}{0.0591} = \dfrac{0.46}{0.0591} \approx 7.79$.

Step 7: Conclusion.
So the answer is X = 0.23 V and $\log K_c = 7.79$. \[ \boxed{0.23,\ 7.79} \]