CBSE Class 12 · 2005 · Delhi Compartment · Redox Reactions and ElectrochemistryShort Answer

A zinc rod is dipped in a 0.1 M solution of ZnSO

_{4}

. The salt is 95% dissociated at this dilution at 298 K. Calculate the electrode potential. (Given:

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

Answer: E ≈ −0.79 V

Solution

Effective [Zn

^{2 +}

0.1 \times 0.95 = 0.095 M

Nernst equation for Zn

^{2 +}

/Zn (

n = 2

E = E^{\circ} + \frac{0.059}{2} lo g [Z n^{2 +}] = - 0.76 + 0.0295 lo g (0.095)

= - 0.76 + 0.0295 \times (- 1.022) = - 0.76 - 0.030

E = - 0.79 V

CBSE Class 12 Board · 2005 · Delhi Compartment · Subjective