Question

Calculate the emf of the following cell at 298 K:
Al(s)/Al3+ (0.15M)//Cu2+(0.025M) /Cu(s)
(Given Eo(Al3+/Al) = -1.66 V, Eo(Cu2+/Cu) = 0.34V, log 0.15 = -0.8239, log 0.025 = -1.6020)

Answer 1

Answer:

1.98V

Explanation:

Here, in this question u have to use nernest equation, and try to solve the log in a non-fractional format and avoid powers for calculating easy logarithms....

Hope u get it...

Answer 2

Given:

• Temperature (T) = 298K
• $Al^{3+} = 0.15M$
• $Cu^{2+} = 0.025M$
• $E^o(Al^{3+}) = -1.66 V$
• $E^o(Cu^{2+}) = 0.34 V$
• log(0.15) = -0.8239
• log(0.025) = -1.6020

To Find:

• EMF of the cell($E_{cell}$).

Solution:

• The formula to find the EMF of any cell is given by, $E_{cell} = E^{o}_{cell} -\frac{0.059}{n} log\frac{(Al^{3+} )^{2} }{(Cu^{2+})^{3} }$   (1)
• We do not know the value of $E^{o}_{cell}$.
• To find  $E^o_{cell}$, we are using the below formula,
• $E^o_{cell} = E^o cathode - E^o anode$ = 0.34 - (-1.66) = 0.34+1.66 = 2 V
• Here $Cu^{2+}$ is the cathode and $Al^{3+}$ is the anode.
• so, $E^o_{cell} = 2V$
• Substituting the values in the first equation we get,
• $E_{cell}=2-\frac{0.059}{6} log \frac{0.15^{2} }{0.025^{3} }$  
• Here n = 6
• $E_{cell} = 2-\frac{0.059}{6}[2log0.15-3log0.025]$  
• $E_{cell} = 2-9.8*10^{-3} [(2*(-0.8239))-(3*(-1.6020))]$
• $E_{cell} = 2-9.8*10^{-3}*[-1.6478+4.806]$
• $E_{cell} = 2-0.0311 = 1.9689 V$

The EMF, E = 1.9689 V