Question

V-13 EXPERIMENT V: VOLTAIC CELLS, AND THE NERNST EQUATION PRELAB ASSIGNMENT (CONTINUED) 5. In aqueous solution Cu* ions are more stable than Cu ions because 2 Cu* (aq) + Cu2+ (aq) + Cu (s) (a) Give an oxidation half-cell reaction and a reduction half-cell reaction that can be combined to form this overall reaction. (0.5 marks) (b) Use Eq. 18 with your answer to (a) to calculate the theoretical K for the reaction. (1 mark)

Equation 18: (didnt mean to attach equation 15 oops)

Answer 1

Ans 5a: 2 Cut (aq) + Cu2+ (aq) + Cu(s) Oxidation Half-cell reaction at anode: Cu+ (aq) + Cu²+ (aq) +12 Reduction Half-cell reaction at cathode: Cut (aq) +1 e - Cu (s) On combining & simplifying we get: Cu* (aq) - Cu2+ (aq) +16 Cu* (aq) +16 + Cu(s) 2 Cut (aq) = Cu2+ (aq) + Cu(s) ------------------------------------

Ans 5b:

0.0592 Eceu = log K n

Let n = number of electrons transferred in the redox reaction (here n = 1)

E^ocell = E^ocathode - E^oanode = E^o_Cu+/Cu - E^o_Cu2+/Cu+

E^ocathode = E^o_Cu+/Cu= standard reduction potential of Cu⁺ +1e ----> Cu = 0.52 (data collected)

E^oanode =E^o_Cu2+/Cu+ =standard reduction potential of Cu²⁺ +1e ----> Cu⁺ = 0.34 (data collected)

E^ocell = E^ocathode - E^oanode = E^o_Cu+/Cu - E^o_Cu2+/Cu+ = 0.52-0.34 = 0.18 V

0.0592 Eceu = log K n

0.18 V = (0.0592/1) logK

K = 1097.84