The measured voltage of the cell Pt(s)| H₂ (1.0 atm)| H⁺(aq)|| Ag⁺(1.0 M)| Ag(s)is 1.02 V at 25°C.Calculate the pH of the solution.

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Quizplus · Accepted Answer

The cell notation indicates that the anode is Pt(s)| H₂ (1.0 atm)| H⁺(aq) and the cathode is Ag⁺(1.0 M)| Ag(s). The half-reactions are: Anode: H⁺(aq) → H+ (aq) + S1U1P4S4P0(aq) + 2e- Cathode: Ag+ (aq) + e- → Ag(s) To balance the half-reactions and cancel the electrons, we need to multiply the anode reaction by 2: 2H⁺(aq) → 2H+ (aq) + S1U1P4S4P0(aq) + 4e- The overall reaction is: 2H⁺(aq) + Ag+ (aq) → 2H+ (aq) + S1U1P4S4P0(aq) + Ag(s) To calculate the standard cell potential, we can use the standard reduction potentials of the half-reactions: E°cell = E°cathode - E°anode E°cell = 0.80 V - (-0.14 V) E°cell = 0.94 V To calculate the cell potential at non-standard conditions, we can use the Nernst equation: Ecell = E°cell - (RT/nF) ln Q where R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin (298 K), n is the number of electrons transferred in the balanced equation (4), F is the Faraday constant (96,485 C/mol), and Q is the reaction quotient: Q = [H+]^2 [S1U1P4S4P0(aq)] / [H⁺(aq)] We can rearrange the Nernst equation to solve for the hydrogen ion concentration: ln Q = (E°cell - Ecell) (nF/RT) [H+]^2 [S1U1P4S4P0(aq)] / [H⁺(aq)] = e^( (E°cell - Ecell) (nF/RT) ) [H+]^2 = [H⁺(aq)] [S1U1P4S4P0(aq)] / [Ag+]^2 e^( (E°cell - Ecell) (nF/RT) ) [H+] = sqrt( [H⁺(aq)] [S1U1P4S4P0(aq)] / [Ag+]^2 e^( (E°cell - Ecell) (nF/RT) ) ) Plugging in the values, we get: [H+] = sqrt( (1.0 M) (1.0 atm) / (1.0 M)^2 e^( -(0.94 V - 1.02 V) (4)(96,485 C/mol) / (8.314 J/mol·K)(298 K) ) ) [H+] = 6.60 × 10^-4 M To calculate the pH, we can use the equation: pH = -log[H+] pH = -log(6.60 × 10^-4) pH = 3.72 Therefore, the answer is (C) 3.72.

The Measured Voltage of the Cell Pt(s)| H2 (1

The Measured Voltage of the Cell Pt(s)| H₂ (1