EMF of Cell

EMF is the voltage developed by any source of electrical energy such as a battery or cell. It represents the potential difference generated between two electrodes in a cell when no current is flowing. The EMF of a cell is measured when no current is flowing through the cell (i.e., in an open circuit). It can be thought of as the maximum potential difference that the cell can deliver.

This Story also Contains

1. EMF of cell
2. Some Solved Examples
3. Summary

EMF of cell

It is the potential difference between the two terminals of the cell when no current is drawn from it. It is measured with the help of potentiometer or vacuum tube voltmeter.

Calculation of the EMF of the Cell
Mathematically, it may be expressed as
$\begin{aligned} & \mathrm{E}_{\text {cell }} \text { or } \mathrm{EMF}=\left[\mathrm{E}_{\text {red }}(\text { cathode })-\mathrm{E}_{\text {red }}(\text { anode })\right] \\ & \mathrm{E}_{\text {elll }}^{\circ} \text { or } \mathrm{EMF}^{\circ} \\ & =\left[\mathrm{E}_{\text {red }}^{\circ}(\text { cathode })-\mathrm{E}_{\text {red }}^{\circ}(\text { anode })\right]\end{aligned}$ Characteristics of cell and cell potential.

• For cell reaction to occur the E_cell should be positive. This can happen only if E_red (cathode) > E_red(anode).
• E^o cell must be positive for a spontaneous reaction.
• It measures free energy change for maximum convertibility of heat into useful work.
• It causes the flow of current from the electrode of the higher E^o value to the lower E^o value.

Difference between EMF and Cell Potential

Recommended topic video on (EMF of cells)

Some Solved Examples

Example.1

1. Which of the following equations connects electrode potential to reaction quotient?

1)Kohlrausch's equation

2) (correct)Nernst equation

3) Ohm's equation

4)Faraday equation

Solution

As we have learned,

Electrode Potential and EMF of Cells -

It is the potential difference between the two terminals of the cell when no current is drawn from it. It is measured with the help of potentiometer or vacuum tube voltmeter.
Nernst equation

$E=E^0-\frac{R T}{n F} \ln Q$

Where E= electrode potential and Q = reaction quotient

Hence, the answer is the option (2).

Example.2

2. If zinc is kept in CuSO₄ solution, copper gets precipitated because the electrode potential of zinc is

1)copper

2) (correct) copper

3)sulphate

4)none

Solution

E_zinc 0=−0.76,E_copper 0=+0.34

The standard electrode potential of zinc < copper and hence Zinc copper from its salt.

Hence, the answer is the option (2).

Example.3

3. Arrange the following in the order of their decreasing electrode oxidizing potential Mg, K, Ba, Ca

1)Ba, Ca, K, Mg

2) (correct)K, Ba, Ca, Mg

3)Ca, Mg, K, Ba

4)Mg, Ca, Ba, K

Solution

As we have learned,

The correct order can be obtained from the electrochemical series.

Hence, the answer is the option (2).

Example.4

4. EMF of a cell in terms of the reduction potential of its left and right electrodes is

1)$E=E_{\text {left }}-E_{\text {right }}$

2)$E=E_{\text {left }}+E_{\text {right }}$

3) (correct)$E=E_{\text {right }}-E_{\text {left }}$

4)$E=-\left(E_{\text {right }}+E_{\text {left }}\right)$

Solution

As we learned from the concept

We know that E cell = Reduction potential of Cathode + Oxidation potential of Anode

= Reduction potential of Cathode - Reduction potential of Anode

= E_right - E_left

Hence, the answer is the option (3).

Example.5

5. If ϕ denotes reduction potential, then which is true?

1) (correct)$E_{c e l l}^c=\phi_{\text {right }}-\phi_{l e f t}$

2)$E_{\text {cell }}^{\circ}=\phi_{\text {left }}+\phi_{\text {right }}$

3)$E_{\text {cell }}^{\circ}=\phi_{\text {left }}-\phi_{\text {right }}$

4)$E_{\text {cell }}^{\circ}=-\left(\phi_{\text {left }}+\phi_{\text {right }}\right)$

Solution

_{$\mathrm{E}_{\text {cell }}^o$} = reduction potential (cathode, right) + Oxidation potential (anode, left)

_{$\mathrm{E}_{\mathrm{cell}}^{\circ}$} = Reduction Potential (right) - Reduction potential (left)

If $\phi$ is reduction potential, then

_{$\mathrm{E}_{\mathrm{cell}}^{\circ}$} = $\varnothing_{\text {right }}-\varnothing_{\text {left }}$

Hence, the answer is the option (1).

Example.6

6. The standard electrode potentials$\left(\mathrm{E}_{\mathrm{M}^{+} \mid \mathrm{M}}^0\right)$ of four metals A, B, C, and D are - 1.2 V, 0.6 V, 0.85 V, and - 0.76 V, respectively. The sequence of deposition of metals on applying potential is :

1) A, C, B, D

2) B, D, C, A

3) (correct)C, B, D, A

4)D, A, B, C

Solution

The higher the standard electrode potential$\mathrm{E}_{\mathrm{M}^{+} \mid \mathrm{M}}^0$ of the metal, the greater its tendency to get reduced and deposited on the electrode.

Order of the given $\mathrm{E}^0$ value $: ~ 0.85 \mathrm{~V}>0.6 \mathrm{~V}>-0.76 \mathrm{~V}>-1.2$

Therfore, The order of deposition will be C, B, D, and A.

Hence, the answer is the option (3).

Summary

The EMF of a battery determines the voltage it can provide to power electronic devices. High EMF batteries, like lithium-ion batteries, are used in smartphones, laptops, and tablets due to their ability to deliver a stable and high voltage while being compact and lightweight. Rechargeable batteries, such as NiMH and Li-ion, rely on their EMF to provide consistent power over many charge-discharge cycles. The EMF of these batteries affects their efficiency and capacity. High EMF cells can store more energy and provide a more stable supply.