Write the Nernst equation and calculate the emf of the following cell at 298 K: $$\mathrm{Cu(s)\,|\,Cu^{2+}(0.130\ M)\,||\,Ag^+(1.00\times10^{-4}\ M)\,|\,Ag(s)}$$ (Given: $E^{\circ}_{\mathrm{Cu^{2+}/Cu}} = +0.34$ V and $E^{\circ}_{\mathrm{Ag^+/Ag}} = +0.80$ V)

CBSE Class 12 · 2004 · All India · Redox Reactions and ElectrochemistryShort Answer

Write the Nernst equation and calculate the emf of the following cell at 298 K:

Cu (s) ∣ C u^{2 +} (0.130 M) ∣∣ A g^{+} (1.00 \times 1 0^{- 4} M) ∣ Ag (s)

(Given:

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

V and

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

Answer: E_cell ≈ 0.25 V

Cell reaction: Cu + 2Ag

^{+} \to

^{2 +}

+ 2Ag,

n = 2

E_{ce l l}^{\circ} = 0.80 - 0.34 = 0.46 V

Nernst equation:

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

Substituting:

E_{ce l l} = 0.46 - 0.0295 lo g \frac{0.130}{( 1 0 ^{- 4} ) ^{2}} = 0.46 - 0.0295 lo g (1.3 \times 1 0^{7})

= 0.46 - 0.0295 \times 7.114 = 0.46 - 0.210

E_{ce l l} = 0.25 V

CBSE Class 12 Board · 2004 · All India · Subjective