Question

A galvanic cell at a temperature of 42 degrees Celcius is powered by the following redox reaction:
3CU2+(aq) + 2Al(s) ----> 3Cu(s) + 2Al3+(aq)
Suppose the cell is prepared with 3.43 M Cu2+n one half-cell and 1.63 M Al3+in the other.
A) Calculate the cell voltage under these conditions. Round your answer to 3 significant digits.

Answer 1

Answer: The cell voltage of the given reaction is 1.86 V

Explanation:

The given chemical equation follows:

$3Cu^{2+}(aq.)+2Al\rightarrow 2Al^{3+}(aq.)+2Au(s)$

Oxidation half reaction: $Al(aq.)\rightarrow Al^{3+}(aq.)+3e^-;E^o_{Al^{3+}/Al}=1.66V$       ( × 2)

Reduction half reaction: $Cu^{2+}(aq.)+2e^-\rightarrow Cu(s);E^o_{Cu^{2+}/Cu}=0.16V$       ( × 3)

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=0.16-(-1.66)=1.82V$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Al^{3+}]^2}{[Cu^{2+}]^3}$

where,

$E_{cell}$ = electrode potential of the cell = ? V

$E^o_{cell}$ = standard electrode potential of the cell = +1.82 V

n = number of electrons exchanged = 2

R = Gas constant = 8.314 J/mol Kl

T = temperature = $42^oC=[42+273]K=315K$

F = Faraday's constant = 96500

$[Al^{3+}]=1.63M$

$[Cu^{2+}]=3.43M$

Putting values in above equation, we get:

$E_{cell}=1.82-\frac{2.303\times 8.314\times 315}{2\times 96500}\times \log(\frac{(1.63)^2}{(3.43)^3})\\\\E_{cell}=1.86V$

Hence, the cell voltage of the given reaction is 1.86 V