(iii) Deviations occur only after equilibrium depending on the value of (iv) ia will be zero (no anodic tendency) only at -ve infinite potential (v) Similarly ic, win be zero (no cahodic tendency) only at +ve infinite potential CONDITION FOR ELECTRODE DEPOSITION = Equilibrium potential for the process considered ∆e = - = - ve cathodic process ∆ ∆e = Cathodic equilibrium potential; ∆e = - = + ve anodic process ∆ ∆e = Anodic equilibrium potential; ∆e (a) Consider the discharge of aq. 1.0 M Cu2+ (reduction) = - = 0.34 < 0 for cathodic; < 0.34V ∆ ∆e ∆- ∆ i.e., the discharge of Cu2+ can occur even at no applied or +ve applied potential but less than 0.34 V w.r.t SHE H.W: Calculate the applied potential for the discharge of 0.01M Cu2+. (Hint: Use Nemst eq for getting ∆e). (b) Consider the discharge of aq. 1.0 M Na+ (reduction) = - = (-2.71) = +2.71 < 0 for cathodic; ∆ ∆e ∆- ∆ < - 2.71 V ∆ i.e., the discharge of Na+ can occur only at -ve applied potential and that too less than - 2.71 V.
(iii) Deviations occur only after equilibrium depending on the value of (iv) ia will be zero (no anodic tendency) only at -ve infinite potential (v) Similarly ic, win be zero (no cahodic tendency) only at +ve infinite potential CONDITION FOR ELECTRODE DEPOSITION ∆e = Equilibrium potential for the process considered = ∆ - ∆e = - ve cathodic process ∆e = Cathodic equilibrium potential; = ∆ - ∆e = + ve anodic process ∆e = Anodic equilibrium potential; (a) Consider the discharge (reduction) of aq. 1.0 M Cu2+ = ∆ - ∆e = ∆- 0.34 < 0 for cathodic; ∆ < 0.34V i.e., the discharge of Cu2+ can occur even at no applied or +ve applied potential but less than 0.34 V w.r.t SHE H.W: Calculate the applied potential for the discharge of 0.01M Cu2+. (Hint: Use Nemst eq for getting ∆e). (b) Consider the discharge (reduction) of aq. 1.0 M Na+ = ∆ - ∆e = ∆- (-2.71) = ∆ +2.71 < 0 for cathodic; ∆ < - 2.71 V i.e., the discharge of Na+ can occur only at -ve applied potential and that too less than - 2.71 V.
03-bv-diss-eta.docx