Question

A zinc rod dipped in n molar solution of ZnSO4 has an electrode potential of -0.56 V. The salt is 98 percent dissociated at room temperature. What is the molarity of the solution? (E°(Zn+2/Zn) = -0.5 V)

(a) 8.44 × 10^-3 M

(b) 9.44 × 10^-4 M

(c) 8.44 × 10^-4 M

(d) 9.44 × 10^-3 M

This question was addressed to me in examination.

This intriguing question originated from Electrochemistry topic in portion Electrochemistry of Chemistry – Class 12

Answer 1

The correct option is (d) 9.44 × 10^-3 M

Easiest explanation: The electrode reaction is Zn^2+ + 2e^– → Zn

The number of electrons transferred, n= 2

Applying Nernst equation, we get E(Zn+2/Zn) = E°(Zn+2/Zn) – \(\frac{0.059}{2}\)log (1 / [Zn^2+])

[Zn^2+] = \(\frac{98}{100}\) × n = 0.98n M

-0.56 = -0.5 – \(\frac{0.059}{2}\)log (1 / 0.98n)

n= 9.44 × 10^-3 M.