Formula
Cell EMF
E°_cell = E°_cathode - E°_anode (both as reduction potentials). E_cell > 0 → spontaneous. Cell notation: anode | anion soln || cation soln | cathode.
-- NCERT Class 12 Chemistry, Ch. 2, p. 34EMF Galvanic
8 MCQs2 revision cards9-step worked example
Source: NCERT Redox Reactions and ElectrochemistryPYQ coverage: NEET 2022Official key: NTA-verifiedLast updated: 25 Sep 2026
EMF Galvanic, explained for NEET
The EMF of a galvanic cell is the potential difference between its two electrodes when no current flows. The single formula that governs this topic:
E°_cell = E°_cathode − E°_anode
Both E° values are standard reduction potentials (NCERT Class 12 Chemistry Chapter 2, page 38). This is the load-bearing detail: you subtract the anode's reduction potential from the cathode's reduction potential. You do NOT flip the sign of either value before subtraction — both stay as reduction potentials.
The trap that costs marks: confusing which electrode is cathode and which is anode. In a galvanic cell, the electrode with the higher reduction potential is the cathode (reduction occurs there), and the lower one is the anode (oxidation occurs there). If E°_cell comes out negative, the reaction is non-spontaneous as written — either you assigned cathode/anode backwards, or the cell genuinely doesn't work in that direction.
How NEET tests this: a question gives you two standard reduction potentials and asks for the cell EMF, or gives the cell notation and asks you to identify cathode/anode and compute E°_cell. The common wrong answer comes from subtracting in the wrong order (cathode − anode vs. anode − cathode) or from accidentally using an oxidation potential without flipping sign.
Watch-out: cell notation convention places the anode on the left and cathode on the right: Anode || Cathode. When you read Zn | Zn²⁺ || Cu²⁺ | Cu, zinc is the anode (left) and copper is the cathode (right). E°_cell = E°_Cu − E°_Zn. If a question gives oxidation potentials, convert them to reduction potentials (flip sign) before applying the formula.
A positive E°_cell confirms the cell reaction is spontaneous under standard conditions.
Can you answer these EMF Galvanic MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
In the standard cell notation Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s), which electrode is the cathode?
Show answer and why every option is right or wrong
Answer: B. By cell notation convention, the electrode on the right of the salt bridge (||) is the cathode. Cu(s) is on the right, so it is the cathode where reduction occurs (NCERT Class 12 Chemistry Chapter 2, page 38).
Why A is wrong: A is wrong because Zn(s) is on the left side of the salt bridge — it is the anode where oxidation occurs, not the cathode.
Why C is wrong: C is wrong because Zn²⁺(aq) is the electrolyte solution at the anode compartment, not an electrode. The cathode is a solid electrode, not an ionic species in solution.
Why D is wrong: D is wrong because Cu²⁺(aq) is the electrolyte in the cathode compartment, not the cathode electrode itself. The cathode is Cu(s), the solid metal electrode where Cu²⁺ ions get reduced.
MCQ 2Easy RecallPractice
The EMF of a galvanic cell under standard conditions is defined as:
Show answer and why every option is right or wrong
Answer: B. The standard cell EMF formula is E°_cell = E°_cathode − E°_anode, where both half-cell potentials are standard reduction potentials (NCERT Class 12 Chemistry Chapter 2, page 38).
Why A is wrong: A is wrong because it reverses the subtraction order. Subtracting cathode from anode gives the negative of the correct EMF. The convention is cathode minus anode, not the reverse.
Why C is wrong: C is wrong because cell EMF is a difference, not a sum. Adding both reduction potentials has no physical meaning in this context. The driving force arises from the difference in tendency to undergo reduction.
Why D is wrong: D is wrong because cell EMF is not a product of the two half-cell potentials. There is no multiplicative relationship between electrode potentials in the EMF formula.
MCQ 3Easy RecallPractice
In a spontaneous galvanic cell, the standard cell EMF (E°_cell) is:
Show answer and why every option is right or wrong
Answer: B. A positive E°_cell indicates a spontaneous cell reaction under standard conditions. If E°_cell were negative, the reaction would be non-spontaneous in the written direction (NCERT Class 12 Chemistry Chapter 2, page 38).
Why A is wrong: A is wrong because E°_cell = 0 means the cell is at equilibrium — no net reaction occurs, so no electrical work is produced. A functioning galvanic cell must have a non-zero EMF.
Why C is wrong: C is wrong because a negative E°_cell means the reverse reaction is spontaneous. A galvanic cell that actually produces electrical work has a positive EMF by definition.
Why D is wrong: D is wrong because for a spontaneous galvanic cell specifically, E°_cell must be positive. A negative value indicates the reaction is non-spontaneous as written. The question asks about a spontaneous cell, so the answer is definite, not ambiguous.
MCQ 4Direct ApplicationPractice
Given E°(Cu²⁺/Cu) = +0.34 V and E°(Zn²⁺/Zn) = −0.76 V, the standard EMF of the Daniell cell (Zn–Cu cell) is:
Show answer and why every option is right or wrong
Answer: C. Cu is the cathode (higher reduction potential) and Zn is the anode. E°_cell = E°_cathode − E°_anode = (+0.34) − (−0.76) = +1.10 V (NCERT Class 12 Chemistry Chapter 2, page 38).
Why A is wrong: A is wrong because −1.10 V results from reversing the subtraction: E°_anode − E°_cathode = (−0.76) − (+0.34) = −1.10 V. The correct order is cathode minus anode.
Why B is wrong: B is wrong because +0.42 V results from subtracting magnitudes without considering signs: 0.76 − 0.34 = 0.42. You must subtract the actual signed values: (+0.34) − (−0.76) = +1.10 V.
Why D is wrong: D is wrong because −0.42 V results from computing 0.34 − 0.76 = −0.42, treating both as positive and then subtracting in the wrong direction. The anode potential is −0.76 V (negative), so the subtraction is (+0.34) − (−0.76).
MCQ 5Direct ApplicationPractice
E°(Ag⁺/Ag) = +0.80 V and E°(Mg²⁺/Mg) = −2.37 V. What is E°_cell for the galvanic cell Mg | Mg²⁺ || Ag⁺ | Ag?
Show answer and why every option is right or wrong
Answer: B. From the cell notation, Mg is the anode (left) and Ag is the cathode (right). E°_cell = E°_cathode − E°_anode = (+0.80) − (−2.37) = +3.17 V (NCERT Class 12 Chemistry Chapter 2, page 38).
Why A is wrong: A is wrong because +1.57 V comes from subtracting magnitudes: 2.37 − 0.80 = 1.57. This ignores the sign of the Mg reduction potential. The correct calculation is (+0.80) − (−2.37) = +3.17 V.
Why C is wrong: C is wrong because −3.17 V results from reversing the subtraction order: E°_anode − E°_cathode = (−2.37) − (+0.80) = −3.17 V. The negative sign would indicate a non-spontaneous reaction, which contradicts the given galvanic cell notation.
Why D is wrong: D is wrong because −1.57 V treats Mg's E° as +2.37 V: 0.80 − 2.37 = −1.57 V. The sign of the Mg reduction potential must be kept: (+0.80) − (−2.37) = +3.17 V.
MCQ 6Direct ApplicationPractice
A galvanic cell has E°_cell = +0.46 V. If the standard reduction potential of the cathode is +0.34 V, the standard reduction potential of the anode is:
Show answer and why every option is right or wrong
Answer: B. E°_cell = E°_cathode − E°_anode. Rearranging: E°_anode = E°_cathode − E°_cell = +0.34 − 0.46 = −0.12 V (NCERT Class 12 Chemistry Chapter 2, page 38).
Why A is wrong: A is wrong because +0.80 V results from adding: 0.34 + 0.46 = 0.80. The formula requires subtraction (E°_cathode − E°_cell), not addition.
Why C is wrong: C is wrong because +0.12 V comes from reversing the subtraction: 0.46 − 0.34 = +0.12. The correct rearrangement is E°_anode = E°_cathode − E°_cell = 0.34 − 0.46 = −0.12 V.
Why D is wrong: D is wrong because −0.80 V results from subtracting as −(0.34 + 0.46) = −0.80. There is no reason to negate the sum. The correct operation is E°_anode = 0.34 − 0.46 = −0.12 V.
MCQ 7Concept TrapPractice
In a galvanic cell, the electrode with the more negative standard reduction potential acts as the:
Show answer and why every option is right or wrong
Answer: B. The electrode with the more negative (lower) reduction potential has a greater tendency to get oxidised. It therefore acts as the anode in a galvanic cell, where oxidation occurs (NCERT Class 12 Chemistry Chapter 2, page 38).
Why A is wrong: A is wrong because the electrode with the more negative reduction potential has LESS tendency to gain electrons (undergo reduction), not more. It is the electrode with the higher reduction potential that acts as cathode.
Why C is wrong: C is wrong because the reasoning is internally contradictory — an electrode with higher tendency to get oxidised acts as the anode (where oxidation occurs), not the cathode. The cathode is the site of reduction.
Why D is wrong: D is wrong because the salt bridge is not an electrode. It is a device (e.g., KCl in agar) that maintains electrical neutrality by allowing ion flow between the two half-cells. Electrodes are the anode and cathode.
MCQ 8Concept TrapPractice
A student computes E°_cell for a proposed galvanic cell and obtains −0.15 V. What does this result indicate?
Show answer and why every option is right or wrong
Answer: B. A negative E°_cell means the forward reaction is non-spontaneous. The reverse reaction, however, would have E°_cell = +0.15 V and is spontaneous. The student's cathode and anode assignments should be swapped for the cell to function as a galvanic cell (NCERT Class 12 Chemistry Chapter 2, page 38).
Why A is wrong: A is wrong because a negative E°_cell means the reaction is non-spontaneous in the forward direction. A galvanic cell producing current requires E°_cell > 0.
Why C is wrong: C is wrong because equilibrium corresponds to E_cell = 0, not E°_cell = −0.15 V. A non-zero EMF (positive or negative) indicates the system is not at equilibrium under those conditions.
Why D is wrong: D is wrong because EMF can be negative — it simply means the forward reaction is non-spontaneous. Negative E°_cell is a valid and meaningful result, not a mathematical error.
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EMF Galvanic: quick recall before you leave
How do you solve a EMF Galvanic question? A worked example
- 1
Given
A galvanic cell is set up with the following half-cells:
• Fe³⁺/Fe²⁺ electrode: E°(Fe³⁺/Fe²⁺) = +0.77 V• Sn⁴⁺/Sn²⁺ electrode: E°(Sn⁴⁺/Sn²⁺) = +0.15 V
Both are standard reduction potentials at 298 K. - 2
Required
Calculate the standard EMF of the galvanic cell and write the cell notation.
- 3
Concept
In a galvanic cell, the electrode with the higher standard reduction potential is the cathode (reduction site), and the lower one is the anode (oxidation site). E°_cell = E°_cathode − E°_anode.
- 4
Formula
E°_cell = E°_cathode − E°_anode
- 5
Substitution
Fe³⁺/Fe²⁺ has the higher reduction potential (+0.77 V) → cathode.
Sn⁴⁺/Sn²⁺ has the lower reduction potential (+0.15 V) → anode.
E°_cell = (+0.77) − (+0.15) - 6
Calculation
E°_cell = 0.77 − 0.15 = 0.62 V
Note: All given values are reduction potentials with two significant figures. No exact-constant adjustment is needed here — both values participate as measured quantities. - 7
Final answer
E°_cell = +0.62 V
Cell notation (anode left, cathode right): Sn²⁺ | Sn⁴⁺ || Fe³⁺ | Fe²⁺ (with Pt electrodes implied for both solution-phase couples)
The positive value confirms the reaction is spontaneous under standard conditions. - 8
Common trap
A frequent error is subtracting in the wrong order: E°_anode − E°_cathode = 0.15 − 0.77 = −0.62 V. The negative sign would wrongly suggest the cell is non-spontaneous. Always subtract anode FROM cathode.
- 9
Similar NEET-style question
Given E°(Cr³⁺/Cr) = −0.74 V and E°(Ni²⁺/Ni) = −0.25 V, calculate E°_cell for the galvanic cell and identify the anode and cathode.
(Answer: Cr is the anode, Ni is the cathode. E°_cell = (−0.25) − (−0.74) = +0.49 V.)
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What to remember before solving EMF Galvanic questions
Which EMF Galvanic formulas do you need for NEET?
Cell EMF
Both as reduction potentials. E°_cell > 0 → spontaneous.
| Symbol | Quantity | SI Unit |
|---|---|---|
| E°_cell | standard cell EMF | V |
| E°_red | reduction potential | V |
Valid when
- Standard conditions (1 M, 1 bar, 298 K)
- Both half-reactions as reductions
More in Redox Reactions and Electrochemistry: 4 exam traps and mistakes · 4 formulas · 2 question patterns from its other lessons.
EMF Galvanic questions from past NEET papers
1 question from NEET 2022. Answers verified against NTA official keys.
NEET 2022
1No, because E°cell = − 2.733 V
2Yes, because E°cell = + 0.287 V
3No, because E°cell = − 0.287 V
4Yes, because E°cell = + 2.733 V
NTA Answer: Option 2(final)
All 21 past-paper questions from Redox Reactions and Electrochemistry →
Sources
NCERT refs: Class 12 Chemistry Chapter 2, p.38
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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