A galvanic cell consists of a Standard Hydrogen Electrode (SHE) as the cathode and another hydrogen electrode ( P_ H ₂ = 1 atm) dipped in an unknown acidic solution as the anode. If the measured cell potential is 0.236 V at 298 K, what is the pH of the unknown solution? ( Given : 2.303 RT F = 0.059 )

Step-by-step solution

The cell can be represented as: Pt (s) | H ₂ (g, 1 atm) | H ⁺ (aq, unknown) || H ⁺ (aq, 1 M) | H ₂ (g, 1 atm) | Pt (s) The cell potential is given by: E_ cell = E_ cathode - E_ anode For the Standard Hydrogen Electrode (cathode): E_ cathode = 0 V For the anode (reduction potential): E_ anode = E^ _ H ⁺/ H ₂ - 0.059 2 P_ H ₂ [ H ⁺]^2 E_ anode = 0 - 0.059 2 1 [ H ⁺]^2 E_ anode = - 0.059 2 (-2 [ H ⁺]) E_ anode = 0.059 [ H ⁺] = -0.059 pH Substituting into the cell potential equation: E_ cell = 0 - (-0.059 pH ) = 0.059