Question

For a beryllium-gold voltaic cell containing Be2+(aq) and Au3+(aq) solutions, do the following. (a) Identify the cathode. (Include states-of-matter under the given conditions in your answer. Type INERT if an inert electrode must be used.)

Answer 1

Answer:  Gold act as cathode.

Explanation:

Standard potential for an electrochemical cell is given by:

$E^0{cell}$ = standard electrode potential =$E^0{cathode}-E^0{anode}$

The $E^0$ values have to be reduction potentials.  

Reduction potentials for given elements

$E^o_{Be^{2+}/Be}=-1.97V$

$E^o_{Au^{3+}/Au}=+1.40V$

The element Be with negative reduction potential will lose electrons undergo oxidation and thus act as anode.The element Au with positive reduction potential will gain electrons undergo reduction and thus acts as cathode.

Cathode : $2Au^{3+}+6e^-\rightarrow 2Au$

Anode : $3Be\rightarrow 3Be^{2+}+6e^-$

$3Be+2Au^{3+}(aq)\rightarrow 2Au+3Be^{2+}(aq)$

Thus gold act as cathode.