A galvanic cell is constructed by coupling a hydrogen gas electrode ( P_ H ₂ = 1 bar ) dipped in an acidic solution of unknown pH with a copper half-cell containing a 0.01 M Cu ²⁺ solution. If the cell potential is 0.458 V at 298 K , the pH of the acidic solution is _________. [Given: E^ _ Cu ²⁺/ Cu = 0.34 V , 2.303RT F = 0.059 V ]

Step-by-step solution

The half-cell reactions are: Anode: H ₂( g ) 2 H ^+( aq ) + 2 e ^- Cathode: Cu ²⁺( aq ) + 2 e ^- Cu ( s ) Overall cell reaction: H ₂( g ) + Cu ²⁺( aq ) 2 H ^+( aq ) + Cu ( s ) Number of electrons transferred, n = 2 . Standard cell potential: E^ _ cell = E^ _ cathode - E^ _ anode = 0.34 V - 0.00 V = 0.34 V Using the Nernst equation: E_ cell = E^ _ cell - 0.059 2 [ H ^+]^2 [ Cu ²⁺] Substitute the given values: 0.458 = 0.34 - 0.0295 [ H ^+]^2 0.01 0.118 = -0.0295 ( [ H ^+]^2 10^2 ) ( [ H ^+]^2 10^2 ) = 0.118 -0.0295 =