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The following data were measured using a nickel electrode as the standard: Cu2+(aq) + 2e- ⟶...

The following data were measured using a nickel electrode as the standard:

Cu2+(aq) + 2e- ⟶ Cu(s) Ni2+(aq) + 2e- ⟶ Ni(s) Fe2+(aq) + 2e- ⟶ Fe(s) Al3+(aq) + 3e- ⟶ Al(s)

The copper and aluminum electrodes are connected in a battery.

A) Which is the anode? Cathode? Why?

B) Which is oxidized? Reduced? Why?

C) What will the battery voltage be?

Eo =0.62V Eo =0.00V Eo =-0.15V Eo =-1.38V

D) Write a balanced net ionic equation for the reaction that takes place.

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Answer #1

More negative reduction potential having stronger reducing agent i.e oxidation takes place at anode and more positive reduction potential having stronger oxidising agent i.e. reduction takes place at cathode.

Among copper and aluminium electrodes

The one with the more negative reduction potential is Aluminium acts as anode and get oxidised .

The one with more positive reduction potential is copper that acts as cathode and get reduced.

Battery voltage, Ecell = Ecathode + Eanode (when sign is changed) = 1.38 V + 0.62 V = +1.97 V

Balanced net ionic equation :

Oxidation half reaction : Al(s) ---> Al+3(aq) + 3e-    x 2 = 2Al(s) ---> 2Al+3(aq) + 6e-

Reduction half reaction : Cu+2(aq) + 2e- ----> Cu(s) x 3 =    3Cu+2(aq) + 6e- ----> 3Cu(s)

Balanced net ionic reaction :    2Al(s) + 3Cu+2(aq) ------> 2Al+3(aq) + 3Cu(s)

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