Answer to Question #120024 in Chemistry for Ronit Chand

Question

Consider the following half-reactions and their standard reduction potential values to answer the following questions.

Cu2+ (aq) + e- Cu+ (aq) E° = 0.15 V
Br2 (l) + 2e- 2Br- (aq) E° = 1.08 V

i) State which reaction occurs at the anode and which at the cathode. [2 marks]
ii) Write the overall cell reaction. [2 marks]
iii) Calculate the value of E°cell. [2 marks]
iv) Calculate the value ΔG° [2 marks]
v) What is the value of Kc at 25°C? [2 marks]

Answer 1

In a voltaic cell, the electrochemical reaction that occurs is the spontaneous reaction. It means that the cell potential must be positive value. The standard cell potential is calculated as the difference between the standard electrode reduction potentials of cathode and anode:

$E^0 = E^0_{cathode}-E^0_{anode}$ .

When one considers the half-reactions:

Cu²⁺ (aq) + e^- $\rightarrow$ Cu⁺ (aq) E° = 0.15 V

Br₂ (l) + 2e^- $\rightarrow$ 2Br^- (aq) E° = 1.08 V,

one can see that the second half-reaction has higher standard electrode potential. This higher standard electrode potential means that the oxidation strength is higher for Br² than for Cu²⁺. In voltaic cell, the reduction reaction occurs on the cathode and the oxidation reaction occurs on the anode:

cathode: Br₂ (l) + 2e^- $\rightarrow$ 2Br^- (aq) E° = 1.08 V,

anode: Cu⁺ (aq) $\rightarrow$ e^-+ Cu²⁺ (aq) E° = -0.15 V.

Therefore, the overall cell reaction is:

2Cu⁺ + Br₂ $\rightarrow$ 2Br^- + 2 Cu²⁺.

The standard cell potential is (the reduction potential of Cu²⁺ must not be multiplied by 2 as it is an intensive quantity):

$E^0 = 1.08 -0.15 = 0.93$ V.

The free energy change is related to the standard cell potential as (using the Faraday constant $F$=96485.3 C/mol, $n$ is the number of the electrons transferred in the reaction):

$∆G^0 = -nFE^0 = -2·96485.3·0.93 = -179.5$ kJ/mol.

The equilibrium constant of the reaction is ($R$ is the ideal gas constant, equal 8.314 J·mol^-1·K^-1):

$K_c = e^{-\frac{∆G}{RT}} = e^{-\frac{-179463}{8.314·298.15}} = 2.77·10^{31}$ .