Question

The Cu²+ and Ag+ ion concentration in a copper-silver electrochemical cell is 0.1M and 0.072M. If E°(Ag/Ag) = 0.8V, E°(Cu²+/Cu) = 0.34V, find out the EMF of the Cell.​

Answer 1

Answer:

The EMF of the given copper-silver electrochemical cell is $0.456\ V$·

Explanation:

Given that,

The concentration of $Cu^{2+}$ $=$ $0.1$ M

The concentration of $Ag^{+}$  $=$ $0.072$ M

The standard reduction potential $E^{0}_{(\ Ag^+/ Ag\ )} \ =\ 0.8 \ V$

The standard reduction potential  $E^{0}_{(\ Cu^2^+/ Cu\ )} \ =\ 0.34 \ V$

        Here, the standard reduction potential of $Ag^+/Ag$ is higher than the standard reduction potential of $Cu^2^+/Cu$· So, the reduction will occur at the silver electrode and the oxidation happens at the copper electrode·

That is, the copper electrode will act as an anode cell and the silver electrode will act as a cathode cell·

Hence, the half-reactions that occur,

At Anode cell,

   $Cu\ --->\ Cu^{2+} \ \ +\ \ 2e^-$

At Cathode cell,

$2\ Ag^{+} \ +\ 2e^---->\ 2Ag$

The overall reaction is,

$Cu\ +\ 2\ Ag^+\ --->\ Cu^{2+}\ +\ 2\ Ag$

Therefore,

The cell representation can be written as,

$Cu\ /\ Cu^{2+}_{\ \ \ (0.1\ M)} \ //\ Ag^+{_(_0_._0_7_2_\ M)}\ /Ag$

The equation to calculate the EMF of the cell is given by Nernst·

So,

The Nernst equation is,

$E_{cell} \ =\ E^0_{cell}\ -\ \frac{0.0591}{n} \ \ log\ \frac{[Cu^2^+]}{[Ag^+]}$

where,

$E_{cell}$  $=$ EMF of the cell

$E^o_{cell}$  $=$ Standard EMF of the cell

n    $=$  number of electrons  $=$  $2\ e^-$

$[Cu^{2+} ]$   $=$ concentration of  $Cu^{2+}$

$[Ag^+]$    $=$ concentration of $Ag^{+}$

From the given standard reduction potential, we can find out the standard reduction potential of the cell·

The equation to calculate the standard reduction potential of the cell is,

$E^o_{cell}$ $=\ E^o_{(cathode)}\ -\ E^o_{(anode)}$

On substituting the values we get,

$E^o_{cell}$   $=$ $0.8\ -\ 0.34$

$E^o_{cell}$   $=$ $0.46$ V

On substituting these values on the Nernst equation we get,

 ⇒  $E_{(cell)} \ =\ 0.46\ -\ \frac{0.0591}{2} \ \ log\ \frac{0.1}{0.072}$

 ⇒  $E_{(cell)} \ =\ 0.46\ -\ \ (0.03 \ \ *\ \ log\ 1.39)$

 ⇒  $E_{(cell)} \ =\ 0.46\ -\ \ (0.03 \ \ *\ \ 0.14)$

 ⇒  $E_{(cell)} \ =\ 0.46\ -\ \ 0.004$

 ⇒  $E_{(cell)} \ =\ 0.456\ V$

Therefore,

The EMF of the given cell is $0.456\ V$·