The equilibrium constant for the following reaction at $$298$$ K is expressed as x×$$10y$$. $$2Fe3++2I$$−→$$2Fe2++I2$$,$$Ecell0=0.235$$ V.The value of y is $$______$$

Δ$$G0=$$−RT $$2.303$$ log $$K-$$ $$nFE0=$$ $$-$$ R T ln K−$$2$$×$$.235$$×$$96500=$$−R×$$298$$× $$2.303$$ log Klog K $$=$$ $$8$$ , therefore $$k=1x108$$

Chemical Equilibrium

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Question

The equilibrium constant for the following reaction at 298 K is expressed as $x \times 10^{y}$ . $2 F e^{3 +} + 2 I^{-} \to 2 F e^{2 +} + I_{2}, E_{c e l l}^{0} = 0.235 V$ .
The value of y is ______

Moderate

Solution

$Δ G^{0} = - R T$ 2.303 log K
- nFE⁰= - R T ln K
$- 2 \times (.235) \times 96500 = - R \times 298 \times$ 2.303 log K
log K = 8 , therefore k=1x10⁸

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