Calculate the emf of the following cell at 298 K: $$\mathrm{Fe(s)\,|\,Fe^{2+}(0.1\ M)\,||\,Ag^+(0.1\ M)\,|\,Ag(s)}$$ Given: $E^{\circ}_{\mathrm{Fe^{2+}/Fe}} = -0.44$ V, $E^{\circ}_{\mathrm{Ag^+/Ag}} = 0.80$ V, R = 8.31 J K$^{-1}$ mol$^{-1}$, 1 F = 96500 C mol$^{-1}$

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

Calculate the emf of the following cell at 298 K:

Fe (s) ∣ F e^{2 +} (0.1 M) ∣∣ A g^{+} (0.1 M) ∣ Ag (s)

Given:

E_{F e^{2 +} /Fe}^{\circ} = - 0.44

E_{A g^{+} /Ag}^{\circ} = 0.80

V, R = 8.31 J K

^{- 1}

mol

^{- 1}

, 1 F = 96500 C mol

^{- 1}

Answer: E_cell ≈ 1.21 V

Cell reaction: Fe + 2Ag

^{+} \to

^{2 +}

+ 2Ag, with

n = 2

E_{ce l l}^{\circ} = 0.80 - (- 0.44) = 1.24 V

Nernst equation:

E_{ce l l} = E_{ce l l}^{\circ} - \frac{0.059}{2} lo g \frac{[ F e ^{2 +} ]}{[ A g ^{+} ] ^{2}}

Substituting:

E_{ce l l} = 1.24 - \frac{0.059}{2} lo g \frac{0.1}{( 0.1 ) ^{2}} = 1.24 - 0.0295 lo g 10

E_{ce l l} = 1.21 V

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