Calculate the potential of the electrochemical cell and determine if it is spontaneous as written at

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

Ecell = -0.825v

Negative results indicate that the electrochemical cell is not spontaneous.

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