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ElectrochemistryNEET Chemistry · Class 12 · NCERT Chapter 2

3 interactive concept widgets for Electrochemistry. Drag any slider, change any number, and watch the formula and the answer update live. Built so you understand how each NEET problem actually works, not just the final number.

Cell EMF calculator

Select any two half-cells from 15 standard electrode potentials. See cell notation, half-reactions, E°cell = E°cathode − E°anode, and whether the cell is spontaneous.

Electrochemistry

Cell EMF calculator

Select any two half-cells from 15 standard electrode potentials. See cell notation, half-reactions, E°cell = E°cathode − E°anode, and whether the cell is spontaneous.

Select two half-cells. The one with the higher E° acts as the cathode (reduction). The one with the lower E° acts as the anode (oxidation). E°cell = E°cathode − E°anode.

Cathode (reduction)

F₂/F⁻

+2.87 V

MnO₄⁻/Mn²⁺

+1.51 V

Cl₂/Cl⁻

+1.36 V

Cr₂O₇²⁻/Cr³⁺

+1.33 V

O₂/H₂O

+1.23 V

Fe³⁺/Fe²⁺

+0.77 V

Cu²⁺/Cu

+0.34 V

H⁺/H₂ (SHE)

+0.00 V

Pb²⁺/Pb

-0.13 V

Ni²⁺/Ni

-0.25 V

Fe²⁺/Fe

-0.44 V

Zn²⁺/Zn

-0.76 V

Al³⁺/Al

-1.66 V

Na⁺/Na

-2.71 V

Li⁺/Li

-3.04 V

Anode (oxidation)

F₂/F⁻

+2.87 V

MnO₄⁻/Mn²⁺

+1.51 V

Cl₂/Cl⁻

+1.36 V

Cr₂O₇²⁻/Cr³⁺

+1.33 V

O₂/H₂O

+1.23 V

Fe³⁺/Fe²⁺

+0.77 V

Cu²⁺/Cu

+0.34 V

H⁺/H₂ (SHE)

+0.00 V

Pb²⁺/Pb

-0.13 V

Ni²⁺/Ni

-0.25 V

Fe²⁺/Fe

-0.44 V

Zn²⁺/Zn

-0.76 V

Al³⁺/Al

-1.66 V

Na⁺/Na

-2.71 V

Li⁺/Li

-3.04 V

Cell notation

Zn|Zn²⁺||Cu²⁺|Cu

Cathode (reduction): Cu²⁺ + 2e⁻ → Cu

Anode (oxidation, reversed): Zn²⁺ + 2e⁻ ← Zn

E°cell = E°cathode − E°anode = (+0.34) − (-0.76) = +1.10 V

Spontaneous (E°cell > 0)

ΔG° = −nFE°cell is negative. The cell does electrical work spontaneously.

NEET tip: higher E° = stronger oxidising agent, lower E° = stronger reducing agent

The metal with the more negative E° is always the anode in a galvanic cell. In NEET, the most tested pair is Zn/Cu (Daniell cell, E°cell = +1.10 V).

Try this

  • Try Zn anode + Cu cathode: the classic Daniell cell with E°cell = 1.10 V.
  • Try Li anode + F₂ cathode: the highest possible cell voltage from this list.
  • Select the same electrode for both anode and cathode to see why a cell cannot work with identical electrodes (unless concentration differs).

Nernst equation explorer

Calculate E from Q (or Q from E) using the Nernst equation. Adjust E°, n, and log Q to see how concentration affects cell potential. Toggle to reverse-calculate Q from a known E.

Electrochemistry

Nernst equation explorer

Calculate E from Q (or Q from E) using the Nernst equation. Adjust E°, n, and log Q to see how concentration affects cell potential. Toggle to reverse-calculate Q from a known E.

The Nernst equation: E = E° − (0.0591 / n) × log Q at 25°C. Use it to find the actual cell potential when concentrations are not 1 M, or to calculate Q when you know E.

Daniell cell (Q=10)
Ag/Ag+ conc. (Q=0.01)
Fe³⁺/Fe²⁺ (Q=0.1)
At equilibrium (E=0)

E°cell (standard EMF)

+1.100 V

n (electrons transferred)

2

log₁₀(Q) reaction quotient

+1.0 (Q = 10^1.01.0e+1)

Q > 1: products accumulate, E decreases

Nernst equation calculation

E = +1.100 − (0.0591 / 2) × 1.0 = +1.1000.0295 = +1.0705 V

Cell is spontaneous

Key relationships (NEET)

Q < K

Reaction proceeds forward, E > 0

Q = K

Equilibrium, E = 0

Q > K

Reaction proceeds reverse, E < 0

E° = (0.0591/n) log K

Standard EMF and equilibrium constant

Try this

  • Set E° = 0 (concentration cell). Any Q not equal to 1 gives a non-zero cell potential.
  • Increase log Q from 0 to 10 and watch E decrease until the cell dies (E ≤ 0).
  • Switch to "Calculate Q from E" and set E = 0 to find the equilibrium constant K.

Faraday's electrolysis calculator

Step-by-step: charge → moles of electrons → moles deposited → mass. Presets for Cu²⁺, Ag⁺, Al³⁺, H₂ gas, O₂ gas. Includes Faraday's second law ratio note.

Electrochemistry

Faraday's electrolysis calculator

Calculate mass deposited and volume of gas evolved during electrolysis. Step-by-step: charge → moles of electrons → moles deposited → mass. Presets for Cu, Ag, Al, H₂, O₂.

Faraday's first law: m = (M × I × t) / (n × F). Enter current, time, molar mass, and n-factor to find the mass deposited during electrolysis.

Cu²⁺ → Cu
Ag⁺ → Ag
Al³⁺ → Al
H₂ gas
O₂ gas

Current (I)

2 A

Time (t)

1800 s (0.50 hr)

Molar mass (M)

63.5 g/mol

n-factor (electrons per ion)

n = 2

Cu²⁺: n=2, Ag⁺: n=1, Al³⁺: n=3, Fe³⁺: n=3, Fe²⁺: n=2

Step 1: Total charge passed

Q = I × t = 2 × 1800

= 3,600 C

Step 2: Moles of electrons

n_e = Q / F = 3,600 / 96500

= 0.0373 mol

Step 3: Moles deposited (n-factor = 2)

moles = n_e / n = 0.0373 / 2

= 0.0187 mol

Step 4: Mass deposited (M = 63.5 g/mol)

m = moles × M = 0.0187 × 63.5

= 1.184 g

Faraday's second law (same charge through two cells)

If the same charge is passed through a CuSO₄ cell (M/n = 63.5/2 = 31.75) and an AgNO₃ cell (M/n = 108/1 = 108), the ratio of masses deposited = 31.75 : 108. NEET frequently tests this as a ratio problem.

Try this

  • Try Cu²⁺ preset: note how n=2 means you need 2× more charge per mole compared to Ag⁺ (n=1).
  • Set current = 1 A and time = 96500 s: exactly 1 Faraday of charge passes. Check how many grams of each substance deposit.
  • Compare H₂ and O₂ gas presets: for the same charge, the mass and volume of H₂ is always less than O₂ because H₂ has M=2 vs O₂ M=32.

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