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JEE Main Chemistry Electrochemistry 2026 JEE Main 2026 (05 April Shift 2)

JEE Main Chemistry Question (2026) — Solution

Question

One half cell in a voltaic cell is constructed by dipping silver rod in AgNO_3 solution of unknown concentration, other half cell is Zn rod dipped in 1 molar solution of ZnSO_4. A voltage of 1.60 V is measured at 298 K for this cell. What is the concentration of Ag^+ ions used in terms of x (x = [Ag^+]) ? E^ _ Zn^ 2+ /Zn = -0.76 V , E^ _ Ag^+/Ag = +0.80 V , 2.303 RT F = 0.059 V

Options

  1. A. 2 3.9
  2. B. 4 5.9
  3. C. 2.9 2
  4. D. 5.9 4

Answer

B. 4 5.9

Step-by-step solution

The cell reaction is Zn(s) + 2Ag^+(aq) Zn^ 2+ (aq) + 2Ag(s) The standard cell potential is given by: E^ _ cell = E^ _ cathode - E^ _ anode = 0.80 - (-0.76) = 1.56 V Using the Nernst equation: E_ cell = E^ _ cell - 0.059 n [Zn^ 2+ ] [Ag^+]^2 Substituting the given values (n = 2, [Zn^ 2+ ] = 1 M , E_ cell = 1.60 V ): 1.60 = 1.56 - 0.059 2 1 x^2 0.04 = - 0.059 2 (-2 x) 0.04 = 0.059 x x = 0.04 0.059 = 4 5.9 Answer: 4 5.9

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