CBSE Class 12 · 2011 · Delhi Compartment · Redox Reactions and ElectrochemistryShort Answer

For the cell

Zn (s) ∣ Z n^{2 +} (2 M) ∣∣ C u^{2 +} (0.5 M) ∣ Cu (s)

a) write the equation for each half-reaction,
b) calculate the cell potential at 25°C.
(Given:

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

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

Answer: (a) Anode: Zn→Zn²⁺+2e⁻; cathode: Cu²⁺+2e⁻→Cu. (b) E_cell ≈ 1.08 V

Solution

(a) Anode (oxidation):

Zn (s) ⟶ Z n^{2 +} (aq) + 2 e^{-}

Cathode (reduction):

C u^{2 +} (aq) + 2 e^{-} ⟶ Cu (s)

(b)

E_{ce l l}^{\circ} = 0.34 - (- 0.76) = 1.10 V

E_{ce l l} = 1.10 - \frac{0.059}{2} lo g \frac{2}{0.5} = 1.10 - 0.0295 lo g 4

= 1.10 - 0.0295 \times 0.602 = 1.10 - 0.018

E_{ce l l} = 1.08 V

CBSE Class 12 Board · 2011 · Delhi Compartment · Subjective