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 MThis question was addressed to me in examination.This intriguing question originated from Electrochemistry topic in portion Electrochemistry of Chemistry – Class 12

A zinc rod dipped in n molar solution of ZnSO4 has an electrode potent

1.	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 MThis question was addressed to me in examination.This intriguing question originated from Electrochemistry topic in portion Electrochemistry of Chemistry – Class 12
Answer» 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.

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