Question

A voltaic cell made of a Cr electrode in a solution of 1.0 M in Cr3+ and a gold electrode in a solution that is 1.0 M in Au3+.

Answer 1

Answer: a) Anode: $Cr\rightarrow Cr^{3+}+3e^-$

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

b) Anode : Cr

Cathode : Au

c) $Au^{3+}+Cr\rightarrow Au+Cr^{3+}$

d) $E_{cell}=2.14V$

Explanation: - 

a) The element Cr 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.

At cathode: $Au{3+}+3e^-\rightarrow Au$

At anode: $Cr\rightarrow Cr^{3+}+3e^-$

b) At cathode which is a positive terminal, reduction occurs which is gain of electrons.

At anode which is a negative terminal, oxidation occurs which is loss of electrons.

Gold acts as cathode ad Chromium acts as anode.

c) Overall balanced equation:

At cathode: $Au{3+}+3e^-\rightarrow Au$     (1)

At anode: $Cr\rightarrow Cr^{3+}+3e^-$        (2)

Adding (1) and (2)

$Au^{3+}+Cr\rightarrow Au+Cr^{3+}$

d)$E^0_(Cr^{3+}/Cr)$= -0.74 V

$E^0_(Au^{3+}/Au)$= 1.40 V  

$E^0{cell}=E^0{cathode}-E^0{anode}=1.40-(-0.74)=2.14V$

Using Nernst equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Au^{3+}]}{[Cr^{3+}]^}$

where,

n = number of electrons in oxidation-reduction reaction = 3

$E^o_{cell}$ = standard electrode potential = 2.14 V

$E_{cell}=2.14-\frac{0.0592}{3}\log \frac{[1.0}{[1.0]}$

$E_{cell}=2.14$

Thus the standard potential for an electrochemical cell with the cell reaction is 2.14 V.