What would be the EMF of the cell in which the following reaction occurs: aligned & Cd ( S )+2 H ⁺ Cd ²⁺+ H _ 2( ~g ) & [ H ⁺ ]=0.02 M E ^0 ( Cd ²⁺ / Cd )=-0.4 ~V , [ Cd ²⁺ ]=0.01 M and partial pressure of H ₂ gas =0.8 ~atm . aligned

Step-by-step solution

The cell reaction is Cd (s) + 2 H ⁺(aq) Cd ²⁺(aq) + H ₂(g) . The standard cell potential E⁰_ cell is calculated as E⁰_ cathode - E⁰_ anode . Given E⁰( H ⁺/ H ₂) = 0.00 V and E⁰( Cd ²⁺/ Cd ) = -0.40 V . E⁰_ cell = 0.00 - (-0.40) = 0.40 V . Using the Nernst equation at 298 K : E_ cell = E⁰_ cell - 0.0591 n [ Cd ²⁺] P_ H ₂ [ H ⁺]² Here n = 2 , [ Cd ²⁺] = 0.01 M , [ H ⁺] = 0.02 M , and P_ H ₂ = 0.8 atm . E_ cell = 0.40 - 0.0591 2 0.01 0.8 (0.02)² E_ cell = 0.40 - 0.02955 0.008 0.0004 E_ cell = 0.40 - 0.02955 (20) Since