The cell potential for the given electrochemical cell at 298 K is 0.89 V . Pt | H ₂ ( g , 1 bar ) | H ^+ ( aq , pH = x) Fe ³⁺ (0.01 M ), Fe ²⁺ (0.1 M ) | Pt The value of x is _________. (Given: E^ _ Fe ³⁺/ Fe ²⁺ = 0.77 V and 2.303RT F = 0.06 V )

Step-by-step solution

The cell reaction is: 1 2 H ₂( g ) + Fe ³⁺( aq ) H ^+( aq ) + Fe ²⁺( aq ) Here, n = 1 . Applying the Nernst equation: E_ cell = E^ _ cell - 0.06 n [ H ^+][ Fe ²⁺] [ Fe ³⁺] Given E_ cell = 0.89 V and E^ _ cell = 0.77 V - 0 V = 0.77 V . Substituting the values: 0.89 = 0.77 - 0.06 1 [ H ^+] 0.1 0.01 0.12 = -0.06 (10 [ H ^+]) (10 [ H ^+]) = -2 10 [ H ^+] = 10⁻² [ H ^+] = 10⁻³ M Since pH = - [ H ^+] , we get pH = 3 . Thus, x = 3 . Answer: 3