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JEE Main 2026 Preparation: Question Papers, Solutions, Mock Tests & Strategy Unacademy » JEE Study Material » Chemistry » Cell Potential

Cell Potential

In this topic, we will learn about Cell potential. We will also learn how a cell potential is generated in a cell, and how it's functioning in the working of the cell and how to calculate it.

Table of Content
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As we know, a cell consists of two electrodes: an anode and a cathode. The Anode is positively charged and the cathode is negatively charged so there is a separation of charge; this charge separation results in generation of a potential difference, this potential difference is called cell potential or voltage. It is also called emf of the cell. This cell potential generates / flows current in a cell.

To study cell potential; first we need to learn some terms related to a cell.

CELL: There are two types of cell: 

Electrochemical cell and Electrolytic cell. In electrolytic cells, we give / apply voltage to the cell. So, the voltage of an electrolytic cell will be the voltage of the applied electricity. But in an electrochemical cell, the voltage is generated in the cell. So: we generally study the cell potential or voltage of the electrochemical cell. Oxidation of the metal electrode takes place at the anode and reduction takes place at the cathode.

Electrode Potential: A cell has two electrodes anode and cathode they are made up of metals so when the circuit is complete the metal at anode starts losing electrons and the metal at cathode starts gaining electrons. As a result of this transfer of electrons, a charge separation occurs at both the electrodes. This charge separation generates potential at both the electrodes. The potential generated at each electrode is called electrode potential. Electrode potential of a half cell can be determined by using a Standard Hydrogen Electrode. Because the half cell potential of SHE is 0V.

Standard Electrode Potential: Standard electrode potential is the electrode potential of a given electrode at a unit concentration, i.e, 1 mole at 1 ATM pressure and 25℃. The standard reduction potential is the standard electrode potential of the cell because the reduction potential is equal to the oxidation potential; but the oxidation potential is negatively charged.

Cell Potential: The difference in potential difference generated at both the electrodes is called cell potential. In general the difference between the potential difference of the potential generated at each half cell is called cell potential.

                        E⁰cell  =  E⁰cathode  –  E⁰anode

                        Ecell > 0 for a spontaneous process

Cell potential is also called the emf of the cell. It is measured in Volts(V).

EMF = Reduction potential of cathode – Reduction potential of anode

Cell Potential is also related to Gibbs free energy. 

                           ∆G = -nEcellF

Calculation of EMF of the cell:

Taking daniell cell for calculation of emf 

For 1 molar concentration the standard electrode potential of half cell reactions

     At cathode                 Cu2+ + 2e– → Cu is equal to E⁰cathode = 0.34 V.

     At anode                     Zn → Zn2+ + 2e– is equal to E⁰anode = -0.76 V.   

The complete cell reaction of the Daniel cell is

Zn + Cu2+ ⇌ Zn2+ + Cu

The emf of the cell is 

  E⁰cell  =  E⁰cathode  –  E⁰anode

  E⁰cell  =  0.34V –  (-0.76)V

  E⁰cell  =  1.1V

Standard Electrode Potential of Some Cells

 

Half Cell Reaction

Standard Electrode Potential

F2(g) + 2e– ⇌ 2F–(aq)

2.87

Al3+(aq) + 3e– ⇌ Al(s)

-1.67

2H+(aq) + 2e– ⇌ H2(g)

0.00

AgBr(s) + e−⇌ Ag(s) + Br−

0.80

H2O2(aq) + 2H+(aq) + 2e– ⇌ 2H2O(l)

1.78

Fe2+(aq) + 2e– ⇌ Fe(s)

-0.44

PbSO4(s) + 2e−⇌ Pb(s) + SO42−

-0.13

 

Nernst Equation for emf of cell:

                                Ecell = Ecell – (RT/nF) ln Q

 Ecell = cell potential 

Eocell = standard state cell potential

R = universal gas constant (8.31 J/mol K)

T = absolute temperature (in Kelvin)

F = Faraday’s constant (96,485 C/mole e–)

n = number of moles of electrons transferred in the cell

Q = reaction quotient

At constant temperature 25℃.

Ecell = Ecell – (0.0257/n) ln Q

or in terms of log10

Ecell = Eocell – (0.0592/n) log Q

 

Conclusion: 

In this topic we studied cell potential  or emf of the cell and its determination with the help of cell potential we can determine how much current an electrochemical cell or battery can produce. Cell potential also helps us in determining which metal electrodes are good for the construction of a cell. 

faq

Frequently Asked Questions

Get answers to the most common queries related to the JEE Examination Preparation.

Calculate the standard electrode potential of the cell reaction-

 2Fe3+ + Fe → 3Fe2+       if two half reactions with E0 values are given...Read full

Calculate emf of the following cell at 25℃.

  Fe∣Fe2+ (0.001M) ∣ H+ (0.01M) ∣ H2 ​(g) (1bar) P...Read full

Calculate ΔrG⁰ and log Kc for the following reaction at 298 K:

2Cr(s) + 3Cd2+(aq) → 2Cr3+(aq) + 3Cd(s)          ...Read full

The cell in which the following reaction occurs:

2Fe+3(aq)  + 2I –(aq)   → 2Fe2 +(aq) + I​2 (s) ...Read full

 2Fe3+ + Fe → 3Fe2+

      if two half reactions with E0 values are given below

  •  Fe3+ + e– → Fe2+ ; E0 = +0.77 V
  • Fe2+ + 2e– → Fe ; E0 = –0.44 V

Solution:- 

For the given reaction –

2Fe3+  +  Fe →  3Fe2+

We know that

 Fe3+ + e– → Fe2+;      E0 Cathode = +0.77 V

Fe2+ + 2e– → Fe;       E0 Anode   = –0.44 V

In this reaction Fe3+ gets reduced and F gets oxidised

For standard electrode potential

        Ecell =  E⁰cathode  –  E⁰anode

             E   =  0.77V – (- 0.44V)

              E = 1.21V

 

  Fe∣Fe2+ (0.001M) ∣ H+ (0.01M) ∣ H2 ​(g) (1bar) Pt(s)

       E0 (Fe2+/Fe) = −0.44V

       E0 (H+ /H2) = 0.00V

Solution: –

 

     For the given cell reaction-

 Fe(s) +2H+(aq)  →  Fe2+(aq) + H2(g)

The standard emf of the cell will be 

Eocell ​= Eo H+/H2  – Eo Fe2+/Fe​

Eocell ​= 0 − (−0.44) = 0.44V

According to Nernst equation; the cell reaction at 25∘C will be 

Eocell = E0cell – {​0.0591 log[Fe2+]/[H+]2}/2

     = 0.44 – {0.0591 log(0.001)/[0.01]2}/2

           = 0.44 −(0.0591 log10)/2

           = 0.44 −0.0591/2

           = 0.44 – 0.02955

Ecell = 0.4105V ≈ 0.41V

2Cr(s) + 3Cd2+(aq) → 2Cr3+(aq) + 3Cd(s) 

                         Given Eocell = + 0.34V, F = 96500 C mol-1

Solution:-

Standard electrode potential of a given cell is

              Ecell =  E⁰cathode  –  E⁰anode

               Ecell =  −0.40−(−0.74)

    Ecell   = 0.34 V

As we know that Gibbs free energy of the reaction is

ΔGo = −nEocell  

ΔGo= −6 × 96500 × 0.34

ΔGo= 196860J/mol

ΔGo=196.86kJ/mol

log Kc =  nEocell / 0.059 

log Kc = 6×0.34/ 0.059

​log Kc= 34.576

Kc=3.76×1034

2Fe+3(aq)  + 2I –(aq)   → 2Fe2 +(aq) + I​2 (s) 

has EOcell ​=0.236V at 298 K, Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.

Solution:- 

The half reactions of the given cell are;

At anode         2Fe3+(aq) + 2e– → 2Fe2+(aq)

At cathode           2I  → I2 + 2e–

No of moles n = 2

ΔGo = −nEocell​     

ΔGo = −2×96500×0.23

ΔGo = −45548CV

ΔGo= −45.55kJ

log Kc = – ΔGo/ 2.303RT

log Kc = – 45.55/ 2.303RT

log Kc = – 45.55/ 0.059

log Kc =  -7.983

 Kc = 9.616 × 107

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