Question

1)A voltaic cell operates at 298 K according to the following reaction:

4 Fe²⁺_(aq) + O_{2 (g)} + 4 H⁺_(aq) → 4 Fe³⁺ _(aq) + 2 H₂O _(l)

What is the emf of this cell when [Fe²⁺] = 6.5908E-4 M, [Fe³⁺] = 0.699 M, pressure O₂ = 0.540 atm and pH = 3.10?

2)A voltaic cell operates at 298 K according to the following reaction:

3 Fe²⁺_(aq) → Fe _(s) + 2 Fe³⁺ _(aq)

What is the emf of this cell when [Fe³⁺] = 0.554 M and [Fe²⁺] = 7.3989E-4 M?

3)A voltaic cell operates at 298 K according to the following reaction:

3 Fe²⁺_(aq) → Fe _(s) + 2 Fe³⁺ _(aq)

What is the ΔG^o (in kJ) of this cell?

4)Write the overall reaction for the RedOx reaction (below) when it occurs in an acidic environment.

KClO_3(aq) + C₁₂H₂₂O_11(s) → KCl _(aq) + CO_{2 (g}

Answer 1

Kclo_3 + C_12 H_22O_11 rightarrow KCl + CO_2 + H_2O rightarrow 8KclO_3 + C_12 H_22 O_11 rightarrow 5 Kcl + 12 CO_2 + 11 H_2O (1) the two half vacations are: 4 Fe^2+ (aq) rightarrow 4 Fe^3+ (aq) + 4e^- - -- E_0 = -0.771 V + O_2 (g) + 4H^+ (aq) + 4e^- rightarrow 2H_2O (l), E_0 = + 1.229 v 4Fe^2+ (aq) + O_2 (g) + 4H^+(aq) rightarrow 4Fe^3+ (aq) + 2H_2O (l), E_0 = +0.455 v if p^H = 3.107 , [H]^+ = 10^-pH = 7.816 times 10^-4 M Note that, [H_2O] does not appear in the Q expression since H_2O is a pure liquid. E_cell = S_2 - (0.059/n) log Q = 0.455 - (0.059/4) log ([Fe^3+ (aq)^4]/[Fe^2+ (aq)]^4) (PO_2) * (H^+ (aq))^4 = 0.459 - (0.0148) log [(0.699)^4/(6.5908 e^-4)^4] times (0.54)(7.816 times 10^-4)^4] = 0.6007 v (2) This question is also similar to the above problem as l solved so find the value of E.