The cell potential for the cell Pt H ₂( g , P atm ) H ^+( aq , pH =3) Ag ^+( aq , 0.01 M ) Ag ( s ) is 0.89 V at 298 K . The value of P is ________. (Nearest integer) (Given: E^ _ Ag ^+/ Ag = 0.80 V and 2.303RT F = 0.06 V )

Step-by-step solution

The half-cell reactions are: Anode: H ₂( g ) 2 H ^+( aq ) + 2 e ^- Cathode: 2 Ag ^+( aq ) + 2 e ^- 2 Ag ( s ) Overall cell reaction: H ₂( g ) + 2 Ag ^+( aq ) 2 H ^+( aq ) + 2 Ag ( s ) The standard cell potential is: E^ _ cell = E^ _ cathode - E^ _ anode = 0.80 - 0 = 0.80 V From the Nernst equation: E_ cell = E^ _ cell - 0.06 2 [ H ^+]^2 P_ H ₂ [ Ag ^+]^2 Given E_ cell = 0.89 V , [ H ^+] = 10⁻³ M (since pH = 3 ), and [ Ag ^+] = 0.01 M = 10⁻² M . Substituting these values: 0.89 = 0.80 - 0.03 (10⁻³)^2 P (10⁻²)^2 0.09