Hg2Cl2(s)+2e−→ 2Hg(l)+2Cl−; Eo=+0.27 VHg22+(aq)→ 2Hg2+(aq)+2e−;Eo=−0.92 V2Hg(l)→Hg2(aq)2++2e−;Eo=−0.79 VHg2Cl2(s)+2e−→ 2Hg(l)+2Cl−(saturated)Cl2(g)+2e−→ 2Cl−(aq);Eo=+1.36V2.303 RT/F=0.06,pressure is at 1 bar. By using the above data, identify among the following which is/are correct?

A) E Hg 2 + / Hg o = + 0 . 92 × 1 + 0 . 79 × 1 2 = + 0 . 855 V B) Hg 2 Cl 2 ( s ) + 2 e − → 2 Hg ( l ) + 2 Cl − ; ΔG ° = − 2 × F × 0 . 27 = − 0 . 54 F 2 Hg ( l ) → 2 Hg 2 + 4 e − ; ΔG ° = − 4 × F × − 0 . 855 = + 3 . 42 F Cl 2 + 2 e − → 2 Cl − ; ΔG ° = − 2 × F × 1 . 36 = − 2 . 72 F Hg 2 Cl 2 ( s ) + C l 2 → 2 Hg 2 + + 4 Cl − ; ΔG ° = + 0 . 16 F − 2 F E cell ° = + 0 . 16 F ⇒ E cell o = − 0 . 08 V D) E Cl − / Hg 2 Cl 2 / Hg o = E Hg 2 2 + / Hg o + 0 . 06 2 log ( K . sp ) Hg 2 Cl 2 + 0 . 27 = + 0 . 79 + 0 . 03 log ( K sp ) Hg 2 Cl 2 K sp of Hg 2 Cl 2 = 10 − 17 . 3 C) E Cl − / Hg 2 Cl 2 / Hg = E Cl − / Hg 2 Cl 2 / Hg o − 0 . 06 2 log [ Cl − ] 2 + 0 . 24 = + 0 . 27 − 0 . 03 log [ Cl − ] 2 − 0 . 03 = 0 . 03 log [ Cl − ] 2 [ Cl − ] = 10 M

Hg 2 Cl 2 ( s ) + Cl 2 ( s ) → 2 Hg 2 + ( aq ) + 4 Cl − ( aq ) E cell o = − 0 . 08 V

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${Hg}_{2} {Cl}_{2} (s) + 2 e^{-} \to_{}^{} 2 Hg (l) + 2 {Cl}^{-}; E^{o} = + 0.27 V$
${Hg}_{2}^{2 +} (aq) \to_{}^{} 2 {Hg}^{2 +} (aq) + 2 e^{-}; E^{o} = - 0.92 V$
$2 Hg (l) \to {Hg}_{2 (aq)}^{2 +} + 2 e^{-}; E^{o} = - 0.79 V$
${Hg}_{2} {Cl}_{2} (s) + 2 e^{-} \to_{}^{} 2 Hg (l) + 2 {Cl}^{-} (saturated)$
${Cl}_{2} (g) + 2 e^{-} \to_{}^{} 2 {Cl}^{-} (aq); E^{o} = + 1.36 V$

$2.303 RT / F = 0.06,$ pressure is at 1 bar. By using the above data, identify among the following which is/are correct?

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${Hg}_{2} {Cl}_{2} (s) + {Cl}_{2} (s) \to_{}^{} 2 {Hg}^{2 +} (aq) + 4 {Cl}^{-} (aq)$

$E_{cell}^{o} = - 0.08 V$

Conc. of KCl in saturated KCl of Calomel electrode saturated with KCl is $\sqrt{10} M$

$E_{{Hg}^{2 +} / Hg}^{o} = + 0.855 V$

$K_{sp} of {Hg}_{2} {Cl}_{2}$ is nearly $10^{- 17.33}$

answer is A, B, C, D.

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Detailed Solution

A) $E_{{Hg}^{2 +} / Hg}^{o} = \frac{+ 0.92 \times 1 + 0.79 \times 1}{2} = + 0.855 V$
B) ${Hg}_{2} {Cl}_{2} (s) + 2 e^{-} \to_{}^{} 2 Hg (l) + 2 {Cl}^{-}; ΔG ° = - 2 \times F \times 0.27 = - 0.54 F 2 Hg (l) \to_{}^{} 2 {Hg}^{2 +} 4 e^{-}; ΔG ° = - 4 \times F \times - 0.855 = + 3.42 F {Cl}_{2} + 2 e^{-} \to_{}^{} 2 {Cl}^{-}; ΔG ° = - 2 \times F \times 1.36 = - 2.72 F {Hg}_{2} {Cl}_{2} (s) + C l_{2} \to_{}^{} 2 {Hg}^{2 +} + 4 {Cl}^{-}; ΔG ° = + 0.16 F - 2 F E_{cell}^{°} = + 0.16 F \Rightarrow E_{cell}^{o} = - 0.08 V$
D) $E_{{Cl}^{-} / {Hg}_{2} {Cl}_{2} / Hg}^{o} = E_{{Hg}_{2}^{2 +} / Hg}^{o} + \frac{0.06}{2} \log {(K . sp)}_{{Hg}_{2} {Cl}_{2}} + 0.27 = + 0.79 + 0.03 \log {(K_{sp})}_{{Hg}_{2} {Cl}_{2}} K_{sp} of {Hg}_{2} {Cl}_{2} = 10^{- 17.3}$
C) $E_{{Cl}^{-} / {Hg}_{2} {Cl}_{2} / Hg} = E_{{Cl}^{-} / {Hg}_{2} {Cl}_{2} / Hg}^{o} - \frac{0.06}{2} \log {[{Cl}^{-}]}^{2} + 0.24 = + 0.27 - 0.03 \log {[{Cl}^{-}]}^{2} - 0.03 = 0.03 \log {[{Cl}^{-}]}^{2} [{Cl}^{-}] = \sqrt{10} M$