For the following cell reaction at 298 K : Zn(s) + 2 Ag ^+(0.1 M ) Zn ²⁺(x M ) + 2 Ag(s) The Gibbs free energy change ( G ) for the reaction is -306.87 kJ mol ⁻¹ . If x = 10^ -y , the value of y is _____. [Given: E^ _ Zn ²⁺/ Zn = -0.76 V , E^ _ Ag ^+/ Ag = 0.80 V , F = 96500 C mol ⁻¹ , 2.303 RT F = 0.06 V ]

Step-by-step solution

G = -nFE_ cell Here, n = 2 . -306.87 10^3 = -2 96500 E_ cell E_ cell = 306870 193000 = 1.59 V Standard cell potential: E^ _ cell = E^ _ cathode - E^ _ anode = 0.80 - (-0.76) = 1.56 V Using Nernst equation: E_ cell = E^ _ cell - 0.06 n [ Zn ²⁺] [ Ag ^+]^2 1.59 = 1.56 - 0.06 2 x (0.1)^2 0.03 = -0.03 x 10⁻² x 10⁻² = -1 x 10⁻² = 10⁻¹ x = 10⁻³ M Comparing with x = 10^ -y , we get y = 3 . Answer: 3