Question

Given $E_{F{e}^{+3}/F{e}^{+2}}^0=+0.76V$ and $E_{I_2/I^{-}}^0=+0.55V.$ 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, $E_{F{e}^{+3}/F{e}^{+2}}^o=+0.76V$ (Cathode)

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

$\therefore E_{Cell}^o=E_C^o-E_A^o=0.76-0.55=0.21$

The complete cell reaction is

$2F{e}^{3+}+2I^{-}\to 2F{e}^{2+}+I_2$

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

$\Rightarrow \log K_C\approx 7$

$\Rightarrow K_C=10^7$