Question

A galvanic cell consists of a metallic. zinc plate immersed in 0.1 M $$Zn{(N{O}_3)}_2$$ solution and metallic plate of lead in 0.02 M $$Pb{(N{O}_3)}_2$$ solution. Calculate the emf
of the cell write the chemical equation for the electrode reactions and represent the
cell. $$(Given,{E^{\circ }}_{Z{n}^{2+},Zn}=-0.76V,{E^{\circ }}_{P{b}^{2+},Pb}=0.13V)$$

Answer 1

Hint: To solve this, you can use the Nernst equation. It is an equation which relates the reduction potential of an electrochemical reaction i.e. half-cell or full-cell reaction to the standard electrode potential. The relation is given as-
$$E_{cell}={E^{\circ }}_{cell}-{\displaystyle \frac{0.597}{n}}\log Q$$
where, Q is the mathematical product of the concentrations of the products of the reaction divided by the mathematical product of the concentrations of the reactants.

Complete step by step answer:
To solve this, firstly let us understand what an emf of a cell is.
In electrochemistry, in a galvanic cell there are two electrodes. We refer to one of them as the anode and other as the cathode. The anode is the negative unit and oxidation takes place at the anode and reduction takes place at the cathode and it is the positive section.
The difference between the electrodes gives us the maximum potential difference between the two cells. This potential is termed as the cell potential and it is given by-
     $$E_{cell}=E_{cathode}-E_{anode}$$

This difference of the potential at these two electrodes calculates the potential across the cell and is termed as EMF of the cell.
Now let us discuss the given question.
Here Zn undergoes oxidation by loss of electrons, thus acting as anode. Lead undergoes reduction by gain of electrons and thus acts as cathode.
     $$Zn+P{b}^{2+}\to Z{n}^{2+}+Pb$$

First, we have to calculate the standard cell potential, It is the potential of the cell under standard state conditions, which is approximated with concentrations of 1 mole per liter (1 M) and pressures of 1 atmosphere at 25 degree Celsius.
     $$\begin{array}{rl}& {E^{\circ }}_{Z{n}^{2+},Zn}=-0.76V\\ & {E^{\circ }}_{P{b}^{2+},Pb}=0.13V\\ & E^{\circ }={E^{\circ }}_{P{b}^{2+},Pb}-{E^{\circ }}_{Z{n}^{2+},Zn}\\ & E^{\circ }=0.13-(-0.76)=0.89\end{array}$$

Now, by using Nernst equation:
$$E_{cell}={E^{\circ }}_{cell}-{\displaystyle \frac{0.597}{n}}\log {\displaystyle \frac{P{b}^{2+}}{Z{n}^{2+}}}$$

where, n is the number of electrons involved in oxidation-reduction reaction which is 2.
$$E_{cell}=0.89-{\displaystyle \frac{0.597}{2}}\log {\displaystyle \frac{0.02}{0.1}}=0.8693$$

We can see from the above calculation that the difference across the two electrodes came out to be 0.8693 which is almost equal to 0.87V.
Therefore, the emf of the cell is 0.87V and this is the required answer.

Note: The value of this electrode potential depends upon certain factors.
- The electrode potential depends upon the temperature. Depending upon the type of reaction (exothermic or endothermic), the equilibrium of the reaction shifts towards left (or right) if we increase (or decrease) the temperature, thus the electrode potential becomes more negative (or positive).
- The electrode potential also depends on the nature of the ions i.e. position in the electrochemical series.
- Increases or decreases in pressure also affects the electrode potential.