Calculate the equilibrium constant for the following reaction at 298 K: $$\mathrm{Cu(s)+Cl_2(g)\longrightarrow CuCl_2(aq)}$$ Given: R = 8.314 J K$^{-1}$ mol$^{-1}$, $E^{\circ}_{\mathrm{Cu^{2+}/Cu}} = 0.34$ V, $E^{\circ}_{\mathrm{\frac{1}{2}Cl_2/Cl^-}} = 1.36$ V, 1 F = 96500 C mol$^{-1}$

CBSE Class 12 · 2005 · All India Compartment · Redox Reactions and ElectrochemistryShort Answer

Calculate the equilibrium constant for the following reaction at 298 K:

Cu (s) + C l_{2} (g) ⟶ CuC l_{2} (aq)

Given: R = 8.314 J K

^{- 1}

mol

^{- 1}

E_{C u^{2 +} /Cu}^{\circ} = 0.34

E_{\frac{1}{2} C l_{2} /C l^{-}}^{\circ} = 1.36

V, 1 F = 96500 C mol

^{- 1}

Answer: K ≈ 4 × 10³⁴

_{2}

is reduced (cathode), Cu oxidised (anode),

n = 2

E_{ce l l}^{\circ} = 1.36 - 0.34 = 1.02 V

At 298 K:

lo g K = \frac{n E _{ce l l}^{\circ}}{0.059} = \frac{2 \times 1.02}{0.059} = 34.58

K \approx 3.8 \times 1 0^{34}

CBSE Class 12 Board · 2005 · All India Compartment · Subjective