Question

Given $E^0_{Fe^{+3}/Fe^{+2}} = +0.76V$ and $E^0_{I_2/I^-} = +0.55V.$ The equilibrium constant for the reaction taking place in galvanic cell consisting of above two electrodes is $\left[\frac{2.303RT}{F}=0.06\right]$

• A. $1 \times 10^7$
• B. $1 \times 10^9$
• C. $3 \times 10^8$
• D. $5 \times 10^{12}$

Answer 1

Answer: (A)

Given, $E_{F e^{+3} / F e^{+2}}^{o}=+0.76 V$ (Cathode)

$E_{I_{2 / I^{-}}}^{o}=+0.55 V$ (Anode)

$\therefore E_{C e l l}^{o}=E_{C}^{o}-E_{A}^{o}=0.76-0.55=0.21$

The complete cell reaction is

$2 Fe ^{3+}+2 I^{-} \rightarrow 2 Fe ^{2+}+I_{2}$

$\therefore E_{\text {cell }}^{o}=\frac{0.059}{2} \log K_{C}$

$\Rightarrow \log K_{C} \approx 7$

$\Rightarrow K_{C}=10^{7}$