For a Daniel cell at 298 K , a plot of E_ cell versus ₁₀ [ Zn ²⁺] [ Cu ²⁺] yields a straight line that intersects the x-axis at 36 . If the concentration of Zn ²⁺ is 10^4 times that of Cu ²⁺ , the cell potential is found to be y 10⁻² V . The value of y is ____. [ Given: 2.303RT F = 0.06 V ]

Step-by-step solution

The cell reaction for a Daniel cell is: Zn(s) + Cu ²⁺( aq ) Zn ²⁺( aq ) + Cu(s) Here, n = 2 . The Nernst equation is: E_ cell = E^ _ cell - 0.06 2 ₁₀ [ Zn ²⁺] [ Cu ²⁺] The x-intercept of the plot occurs when E_ cell = 0 . 0 = E^ _ cell - 0.03 36 E^ _ cell = 1.08 V When the concentration of Zn ²⁺ is 10^4 times that of Cu ²⁺ : [ Zn ²⁺] [ Cu ²⁺] = 10^4 E_ cell = 1.08 - 0.03 ₁₀ (10^4) E_ cell = 1.08 - 0.03 4 = 1.08 - 0.12 = 0.96 V The cell potential is 96 10⁻² V , so y = 96 . Answer: 96