Question

You are supplied with the following substances:
Copper rod, Zinc rod, salt bridge, two glass beakers, a piece of wire, 1 M $CuS{O_4}$ solution, 1M $ZnS{O_4}$ solution
(a) represent the cell made using the above materials.
(b) (1) Write the Nernst equation for the above cell
(2) Calculate the standard EMF of the cell if
$E_{(Z{n^{2 + }}/Zn)}^0 = - 0.76V$
$E_{(C{u^{2 + }}/Cu)}^0 = + 0.34V$

Answer 1

In the two beakers, two electrodes are dipped one is copper rod and other is zinc rod containing two solutions copper sulphate and zinc sulphate each of 1 molar. The two solutions are linked by a salt bridge. The copper rod behaves as a cathode and zinc rod behaves as anode. In anode oxidation takes place and in cathode reduction takes place.

Complete step by step answer:
Given,
Molarity of copper sulphate solution is 1 M.
Molarity of zinc sulphate solution is 1 M.
Here, the copper rod behaves as a cathode where oxidation is taking place and a zinc rod behaves as anode where reduction is taking place.
The oxidation reaction is defined as the loss of electrons and reduction reaction is defined as the gain of electrons.
The reaction taking place at anode is shown below.
$Zn \to Z{n^{2 + }} + 2{e^ - }E_{ox}^0 = - 0.76V$
The reaction taking place at cathode is shown below.
$C{u^{2 + }} + 2{e^ - } \to Cu\;\;E_{red}^0 = 0.34V$
(a) For the given reaction, the cell is represented as shown below.
$Zn|ZnS{O_4}||CuS{O_4}|Cu$
(b)(1) The Nernst equation is given as shown below.
$E = {E^0} - \dfrac{{RT}}{{nf}}\log \dfrac{{[Oxidation]}}{{[\operatorname{Re} duction]}}$
Where,
E is the cell potential.
${E^0}$ is the cell potential.
R is the ideal gas constant
T is the temperature
n is the number of moles
F is the Faraday's constant
The Nernst equation for the given reaction is shown below.
$E = {E^0} - \dfrac{{RT}}{{nf}}\log \dfrac{{[ZnS{O_4}]}}{{[CuS{O_4}]}}$
(2)
The cell potential is calculated as shown below.
$E_{cell}^0 = E_{reduction}^0 + E_{Oxidation}^0$
To calculate the cell potential, substitute the value in the above equation.
$\Rightarrow E_{cell}^0 = 0.34 + ( - 0.76)$
$\Rightarrow E_{cell}^0 = - 0.42V$
Therefore, the correct option is C.

Note:
The salt bridges are usually made up of potassium nitrate, potassium chloride, sodium chloride which is used to maintain the charge balance as the movement of electrons takes place from one electrode to another.