Hello. This question is incomplete. The full question is:
Calculate the potential of the electrochemical cell and determine if it is spontaneous as written at 25 ∘C . Cu(s) || Cu2+(0.14 M) ‖‖ Fe2+(0.0040 M) || Fe(s) E∘Cu2+/Cu=0.339 VE∘Fe2+/Fe=−0.440 V Ecell= V Is the electrochemical cell spontaneous or not spontaneous as written at 25 ∘C ? not spontaneous spontaneous
Answer:
Ecell = -0.825v
Negative results indicate that the electrochemical cell is not spontaneous.
Step-by-step explanation:
You will find the result by doing the following calculation:
Cu2 + = 0.14M
Fe2 + = 0.0040M
Eº Cu2 + / Cu = 0.339V
Eº Fe2 + / Fe = -0.440V
Electron change ............ n = 2
Cu // Cu2 + (0.14M) // Fe2 + (0.0040M) / Fe (s)
Cu + Fe2 + ----> Fe (s) + Cu2 + (aq)
You should now use the formula =
Eºcell = Eº anode - Eº cathode ------> Eº = Eº (Fe2 + / Fe) - Eº (Cu2 + / Cu)
Eº Cell = -0.440 - 0.339 = -0.779V
E cell = EºCell - (0.0591 / n) log (Cu2 + / Fe2 +)
E cell = -0.779 - (0.0591 / 2) log (0.14 / 0.0040)
Ecell = -0.825