Question

A galvanic cell at a temperature of 42 degrees Celcius is powered by the following redox reaction:

Answer 1

Answer: The cell voltage of the given cell is 2.01

Explanation:

The given chemical cell equation follows:

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

Oxidation half reaction: $2Al(s)\rightarrow 2Al^{3+}(aq,1.63M)+3e^-;E^o_{Al^{3+}/Al}=-1.66V$       ( × 2)

Reduction half reaction: $3Cu^{2+}(aq,3.43M)+2e^-\rightarrow 3Cu(s);E^o_{Cu^{2+}/Cu}=0.34V$      ( × 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:

$E^o_{cell}=0.34-(-1.66)=2.00V$

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 = +2.00 V

R = Gas constant = 8.314 J/mol.K

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

F = Faraday's constant = 96500

n = number of electrons exchanged = 6

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

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

Putting values in above equation, we get:

$E_{cell}=2.00-\frac{2.303\times 8.314\times 315}{6\times 96500}\times \log(\frac{(1.63)^2}{(3.43)^3})$

$E_{cell}=2.01$

Hence, the cell voltage of the given cell is 2.01