Question

Calculate the values of Delta G and Ecell for the following reaction, label the half reactions as anode and cathode.

2Ag(s) + Cu²⁺ (aq) > 2 Ag⁺ (aq) + Cu(s)

* My main question lies in which is cathode and anode...it seems that the Ag half reaction (2Ag(s) > 2Ag⁺ +2e-) signifies Ag(s) loosing electrons causing the positive reactant, this leads me to think loss of electrons/this is the oxidation half reaction, therefore Ag(s) is the reduction agent and anode with a given appendix value of 0.800V, BUT when I look at my professors notes he signifies that the Cu²⁺(aq) half reaction (Cu²⁺  + 2e- > Cu(s)) is the anode with a appendix value of 0.342V.

To me the Cu²⁺ gains electrons and is the reduction reaction/cathode - why am I wrong, is it because the Ag value is the more positive of the two? and if this is so... When am I to determine the cathode/anode label with the "more positive" logic?

Answer 1

Solution :-

2Ag(s) + Cu²⁺ (aq) > 2 Ag⁺ (aq) + Cu(s)

For the given reaction

Ag(s) is oxidized and Cu^2+(aq) is reduced

Therefore the half reactions are as follows

2Ag(s)   ---------- > 2 Ag^+(aq) +2 e-                             oxidation (anode)         Eo = 0.80 V

Cu^2+(aq) + 2e- ------- > Cu(s)                                     reduction (cathode)      Eo = 0.342 V

Now lets calculate the Eo cell

Formula

Eo cell = E cathode – E anode

             = 0.342 V – 0.80 V

              = -0.458 V

Eo cell = - 458 V

Now lets calculate the delta G of the reaction

Delta Go = - n* F* Eocell

n= number of electrons transferred

F= (faradays constant 96485 C)

Now lets put the values in the formula

Delta G = - 2*96485 C per mol *(-0.458 V)

             = 88380 J /mol

88380 J * 1 kJ/1000 J = 88.380 kJ

Therefore the delta G o of the reaction is 88.380 kJ