Consider the following electrochemical cell at 298 K : Pt(s) H ₂( g , 1 atm ) H ^+( aq , pH = x) Sn ⁴⁺( aq , 0.1 M ), Sn ²⁺( aq , 0.1 M ) Pt(s) If the cell potential is 0.27 V , the value of x is _______. (Given: E^ _ Sn ⁴⁺/ Sn ²⁺ = 0.15 V , 2.303RT F = 0.06 V )

Step-by-step solution

The cell reaction involves the oxidation of hydrogen gas at the anode and the reduction of Sn ⁴⁺ at the cathode. Anode: H ₂( g ) 2 H ^+( aq ) + 2 e ^- Cathode: Sn ⁴⁺( aq ) + 2 e ^- Sn ²⁺( aq ) Overall reaction: H ₂( g ) + Sn ⁴⁺( aq ) 2 H ^+( aq ) + Sn ²⁺( aq ) The number of electrons transferred, n = 2 . The standard cell potential is: E^ _ cell = E^ _ cathode - E^ _ anode = 0.15 V - 0 V = 0.15 V Applying the Nernst equation: E_ cell = E^ _ cell - 0.06 2 [ H ^+]^2 [ Sn ²⁺] P_ H ₂ [ Sn ⁴⁺] Given E_ cell = 0.27 V , [