One half cell in a voltaic cell is constructed by dipping silver rod in AgNO₃ solution of unknown concentration, other half cell is Zn rod dipped in 1 molar solution of ZnSO₄ . 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^ _ Zn²⁺/Zn = -0.76 V , E^ _ Ag^+/Ag = +0.80 V , 2.303 RT F = 0.059 V

Step-by-step solution

The cell reaction is Zn(s) + 2Ag^+(aq) Zn²⁺(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 [Zn²⁺] [Ag^+]^2 Substituting the given values ( n = 2 , [Zn²⁺] = 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