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.
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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 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).
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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.
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.
E°cell (standard EMF)
+1.100 V
n (electrons transferred)
2
log₁₀(Q) reaction quotient
+1.0 (Q = 10^1.0 ≈ 1.0e+1)
Q > 1: products accumulate, E decreases
Nernst equation calculation
E = +1.100 − (0.0591 / 2) × 1.0 = +1.100 − 0.0295 = +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
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.
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.
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.
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