Consider the cell Pt | H ₂( ~g )(1 bar ) | H ⁺( aq ) | Cu ²⁺( aq ) Cu At 298 K , the emf of the cell is 0.31 V . The pH of the acidic solution is 3 . The concentration of Cu ²⁺ is:

Step-by-step solution

The cell reaction is H ₂(g) + Cu ²⁺(aq) 2 H ⁺(aq) + Cu (s) . The standard cell potential E^ _ cell = E^ _ cathode - E^ _ anode = E^ _ Cu ²⁺/ Cu - E^ _ H ⁺/ H ₂ = 0.34 V - 0.00 V = 0.34 V . Using the Nernst equation at 298 K: E_ cell = E^ _ cell - 0.0591 n Q Here n = 2 , Q = [ H ⁺]^2 P_ H ₂ [ Cu ²⁺] . Given pH = 3 , so [ H ⁺] = 10⁻³ M . P_ H ₂ = 1 bar . 0.31 = 0.34 - 0.0591 2 ( (10⁻³)^2 1 [ Cu ²⁺] ) -0.03 = -0.02955 ( 10⁻⁶ [ Cu ²⁺] ) Approximating 0.0591/2 0.03 , we get 1 ( 10⁻⁶ [ Cu ²⁺] ) . Therefore, 10^1 = 10⁻⁶ [