Question

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

Answer 1

Answer:

The cell voltage of the given cell is 2.01 V

Explanation:

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

Oxidation half reaction:

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

Reduction half reaction:

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

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

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

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

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

$[Cu^{2+}]=3.43M[Al^{3+}]=1.63M$

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

$E_{cell}=2.01$

Thus, the cell voltage of the given cell is 2.01

Answer 2

Answer:

1.99V

Explanation:

Balanced redox reaction equation:

3CU2+(aq) + 2Al(s) ------> 3Cu(s) + 2Al3+(aq)

E°cell= E°cathode- E°anode

E°cell= 0.34-(-1.66)

E°cell= 2.0V

From Nernst equation:

E= E°cell - 0.0592/n logQ

E= 2.0 - 0.0592/6 log [3.43]/[1.63]

E= 2.0- 0.0032

E= 1.99V