A galvanic cell consists of a standard zinc half-cell and a hydrogen half-cell. The cell is represented as: Zn(s) Zn ²⁺(0.1 M ) H ⁺( aq ) H ₂(1 bar ) Pt If the measured emf of the cell is 0.642 V at 298 K , what is the pH of the solution in the hydrogen half-cell? (Given: E^ _ Zn ²⁺/ Zn = -0.76 V and 2.303 RT F = 0.059 V at 298 K )

Step-by-step solution

The cell reaction is: Zn(s) + 2 H ⁺( aq ) Zn ²⁺( aq ) + H ₂( g ) The standard cell potential is: E^ _ cell = E^ _ cathode - E^ _ anode = 0.00 - (-0.76) = 0.76 V The number of electrons transferred, n = 2 . Using the Nernst equation: E_ cell = E^ _ cell - 0.059 n [ Zn ²⁺]P_ H ₂ [ H ⁺]² Substitute the given values: 0.642 = 0.76 - 0.059 2 0.1 1 [ H ⁺]² 0.642 = 0.76 - 0.0295 10⁻¹ [ H ⁺]² 0.76 - 0.642 = 0.0295 ( 10⁻¹ - [ H ⁺]²) 0.118 = 0.0295 (-1 - 2 [ H ⁺]) Since pH = - [ H ⁺] , we can write -2 [ H ⁺] = 2 pH : 0.118 =