Find the value of equilibrium constant ( Kc ) for the chemical equation: Hg22 + arrow Hg+Hg2 + . The following data has been given: Hg22 ++2e arrow 2Hg,E0=0.789V &Hg2 ++2e arrow Hg,E0=0.854V

Question

Find the value of equilibrium constant ( Kc ) for the chemical equation: Hg22 + arrow Hg+Hg2 + . The following data has been given: Hg22 ++2e arrow 2Hg,E0=0.789V &Hg2 ++2e arrow Hg,E0=0.854V (A) 3.13× 10- 3 (B) 3.13× 10- 4 (C) 6.26× 10- 3 (D) 6.26× 10- 4

Tardigrade Admin · Accepted Answer

Hg22 ++2e- arrow 2Hg,0.789 Volt Hg arrow Hg2 ++2e-,-0.854 Volt Hg2 star arrow Hg+Hg2 +,-0.065 Volt Δ G=-2× (.-0.065.)× 96500=-8.314× 298lnKe q Ke q .=6.3× 10- 3

Q. Find the value of equilibrium constant ( $Kc$ ) for the chemical equation: $H g_{2}^{2 +} \to H g + H g^{2 +}$ .
The following data has been given: $H g_{2}^{2 +} + 2 e \to 2 H g, E^{0} = 0.789 V$ $& H g^{2 +} + 2 e \to H g, E^{0} = 0.854 V$

$3.13 \times 1 0^{- 3}$

$3.13 \times 1 0^{- 4}$

$6.26 \times 1 0^{- 3}$

$6.26 \times 1 0^{- 4}$

$H g_{2}^{2 +} + 2 e^{-} \to 2 H g, 0.789$ Volt
$H g \to H g^{2 +} + 2 e^{-}, - 0.854$ Volt
$H g_{2}^{⋆} \to H g + H g^{2 +}, - 0.065$ Volt
$Δ G = - 2 \times (- 0.065) \times 96500 = - 8.314 \times 298 l n K_{e q}$
$K_{e q .} = 6.3 \times 1 0^{- 3}$

Q. Find the value of equilibrium constant ( Kc ) for the chemical equation: Hg22+​→Hg+Hg2+ . The following data has been given: Hg22+​+2e→2Hg,E0=0.789V &Hg2++2e→Hg,E0=0.854V

3.13×10−3

3.13×10−4

6.26×10−3

6.26×10−4