For the reaction, Hg2Cl2(s) + H2(g) → 2Hg(liq) + 2HCl(aq), the correct representation of the cell and the thermodynamic properties ΔG, ΔH and ΔS at 298 K respectively, are (Given: E298 = 0.2684 V and temperature coefficient = 3 × 10-4 VK-1)
A. Pt|H2(g, 1 atm)|HCl(aq)|Hg2Cl2(g)|Hg(liq) ΔG = -51.8 kJ mol-1, ΔH = -69 kJ mol-1, ΔS = -58 JK-1 mol-1
B. Pt|H2(g, 1 atm)|HCl(aq)|Hg2Cl2(g)|Hg(liq) ΔG = -25.9 kJ mol-1, ΔH = -34.5 kJ mol-1, ΔS = -29 JK-1 mol-1
C. Hg(liq)|H2Cl2(g)|HCl(aq)|H2(g, 1 atm)|Pt ΔG = -51.8 kJ mol-1, ΔH = -69 kJ mol-1, ΔS = -58 JK-1 mol-1
D. Hg(liq)|H2Cl2(g)|HCl(aq)|H2(g, 1 atm)|Pt ΔG = 51.8 kJ mol-1, ΔH = 69 kJ mol-1, ΔS = 58 JK-1 mol-1
Answer: Option A
This Question Belongs to Engineering Chemistry >> Electrochemistry
A. $$2.303\frac{{RT}}{F}$$
B. $$ - 2.303\frac{{RT}}{F}$$
C. $$E_{{M^ + }|m}^ \circ + 2.303\frac{{RT}}{F}$$
D. $$E_{{M^ + }|m}^ \circ - 2.303\frac{{RT}}{F}$$
Given that E°(Fe3+, Fe) = -0.4 V and E°(Fe2+, Fe) = -0.44 V, the value of E°(Fe3+, Fe2+) is
A. 0.76V
B. -0.40 V
C. -0.76 V
D. 0.40 V
A. decrease by a factor of 2
B. increase by a factor of 2
C. decrease by a factor of 4
D. increase by a factor of 4
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