Calculate the cell emf at 25°C for the following cell: $$\mathrm{Zn(s)\,|\,Zn^{2+}(0.1\ M)\,||\,Pb^{2+}(0.02\ M)\,|\,Pb(s)}$$ (Given $E^{\circ}_{\mathrm{Zn^{2+}/Zn}} = -0.76$ V, $E^{\circ}_{\mathrm{Pb^{2+}/Pb}} = -0.13$ V, 1 F = 96500 C mol$^{-1}$) Calculate the maximum work that can be accomplished by the operation of this cell.

CBSE Class 12 · 2003 · Delhi · Redox Reactions and ElectrochemistryShort Answer

Calculate the cell emf at 25°C for the following cell:

Zn (s) ∣ Z n^{2 +} (0.1 M) ∣∣ P b^{2 +} (0.02 M) ∣ Pb (s)

(Given

E_{Z n^{2 +} /Zn}^{\circ} = - 0.76

E_{P b^{2 +} /Pb}^{\circ} = - 0.13

V, 1 F = 96500 C mol

^{- 1}

)
Calculate the maximum work that can be accomplished by the operation of this cell.

Answer: E_cell ≈ 0.609 V; W_max ≈ 117.6 kJ mol⁻¹

E_{ce l l}^{\circ} = - 0.13 - (- 0.76) = 0.63 V, n = 2

Nernst equation:

E_{ce l l} = 0.63 - \frac{0.059}{2} lo g \frac{0.1}{0.02} = 0.63 - 0.0295 lo g 5

= 0.63 - 0.0295 \times 0.699 = 0.63 - 0.021

E_{ce l l} = 0.609 V

Maximum work:

W_{ma x} = n F E_{ce l l} = 2 \times 96500 \times 0.609

W_{ma x} = 1.18 \times 1 0^{5} J = 117.6 kJ mo l^{- 1}

CBSE Class 12 Board · 2003 · Delhi · Subjective