Given: Hg 22++2 e- longrightarrow 2 Hg ; E°=0.789 V Hg 2++2 e- longrightarrow Hg ; E°=+0.854 V Then equilibrium constant for Hg 22+ longrightarrow Hg + Hg 2+ is:

Question

Given: Hg 22++2 e- longrightarrow 2 Hg ; E°=0.789 V Hg 2++2 e- longrightarrow Hg ; E°=+0.854 V Then equilibrium constant for Hg 22+ longrightarrow Hg + Hg 2+ is: (A) 3.13 × 10-3 (B) 6.26 × 10-3 (C) 3.13 × 10-4 (D) 6.26 × 10-4

Tardigrade Admin · Accepted Answer

Hg 22++2 e- longrightarrow 2 Hg ; E°=0.789 V Hg 2++2 e- longrightarrow Hg ; E°=+0.854 V Hg 22+ longrightarrow Hg + Hg 2+ E°=+0.789-0.854=-0.065 V K= antilog [(n E°/0.059)]= antilog [(2 ×(-0.065)/0.059)]=6.26 × 10-3

Q. Given:
$H g_{2}^{2 +} + 2 e^{-} ⟶ 2 H g; E^{\circ} = 0.789 V$
$H g^{2 +} + 2 e^{-} ⟶ H g; E^{\circ} = + 0.854 V$
Then equilibrium constant for
$H g_{2}^{2 +} ⟶ H g + H g^{2 +}$ is:

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

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

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

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

Q. Given: Hg22+​+2e−⟶2Hg;E∘=0.789V Hg2++2e−⟶Hg;E∘=+0.854V Then equilibrium constant for Hg22+​⟶Hg+Hg2+ is:

3.13×10−3

6.26×10−3

3.13×10−4

6.26×10−4

Hg22+​+2e−⟶2Hg;E∘=0.789V Hg2++2e−⟶Hg;E∘=+0.854V Hg22+​⟶Hg+Hg2+E∘=+0.789−0.854=−0.065V K= antilog [0.059nE∘​]= antilog [0.0592×(−0.065)​]=6.26×10−3

Q. Given:
$H g_{2}^{2 +} + 2 e^{-} ⟶ 2 H g; E^{\circ} = 0.789 V$
$H g^{2 +} + 2 e^{-} ⟶ H g; E^{\circ} = + 0.854 V$
Then equilibrium constant for
$H g_{2}^{2 +} ⟶ H g + H g^{2 +}$ is:

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

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

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

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