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Class 12 Chemistry Chapter 2: Electrochemistry MCQs with answers

10 multiple-choice questions on Electrochemistry, each with its answer and the reason, written from the NCERT chapter. For the CBSE board, JEE Main and NEET, which are set from the same NCERT chapter. Open a question's solution to check your working; then take the free test to find your next chapter.

Question 1
For the galvanic cell Mg(s) | Mg²⁺(aq) || Pb²⁺(aq) | Pb(s), which statement is correct?
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Answer: (c) Magnesium is the anode and is the negative electrode.

The anode is written on the left: magnesium is oxidised there and is negative. Electrons go through the wire from Mg to Pb, so current goes from Pb to Mg.

Question 2
In the cell notation Al(s) | Al³⁺(aq) || Sn²⁺(aq) | Sn(s), the double vertical line stands for
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Answer: (c) a salt bridge

A single line separates a metal from its electrolyte; the double line is the salt bridge joining the two half-cells.

Question 3
In the standard hydrogen electrode, hydrogen gas and hydrogen ions are kept at
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Answer: (a) H₂ at 1 bar and H⁺ at 1 M, over platinum coated with platinum black

The SHE is platinised platinum in an acidic solution with both forms of hydrogen at unity: H₂ at 1 bar and H⁺ at 1 molar.

Question 4
Given E°/V: Na⁺|Na −2.71, Mg²⁺|Mg −2.36, Al³⁺|Al −1.66, Zn²⁺|Zn −0.76. The strongest reducing agent among the four metals in aqueous solution is
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Answer: (b) Na

The lower the standard electrode potential, the stronger the metal as a reducing agent; sodium has the lowest E° of the four.

Question 5
A clean nickel strip (E° Ni²⁺|Ni = −0.25 V) is dipped, one at a time, into 1 M solutions of Pb²⁺ (E° −0.13 V), Fe²⁺ (−0.44 V), Zn²⁺ (−0.76 V) and Mg²⁺ (−2.36 V). A metal is deposited on the strip from the solution of
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Answer: (d) Pb²⁺ only

Ni reduces an ion only if E°cell = E°(ion) − (−0.25) is positive: true for Pb²⁺ (+0.12 V), false for Fe²⁺, Zn²⁺ and Mg²⁺.

Question 6
Given E°(Al³⁺|Al) = −1.66 V and E°(Ni²⁺|Ni) = −0.25 V, the standard emf of the cell Al(s) | Al³⁺(aq) || Ni²⁺(aq) | Ni(s) is
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Answer: (b) 1.41 V

E°cell = E°right − E°left = −0.25 − (−1.66) = 1.41 V. E° is intensive: the factor 2 or 3 in the balanced equation does not multiply it.

Question 7
Given E°/V: Ag⁺|Ag 0.80, Cu²⁺|Cu 0.34, Ni²⁺|Ni −0.25, Au³⁺|Au 1.40. Which metal can liberate hydrogen from 1 M HCl under standard conditions?
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Answer: (d) Ni

Only a metal with a negative E° reduces H⁺ (E° = 0.00 V) to H₂; of the four, only nickel has a negative E°.

Question 8
In the electrode Pt(s) | Br₂(aq) | Br⁻(aq), platinum
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Answer: (b) takes no part in the reaction but provides a surface for electron transfer

Platinum is an inert electrode: it does not react, but provides the surface for the Br₂/Br⁻ change and conducts electrons.

Question 9
At 298 K, the electrode potential of a copper rod in 0.010 M CuSO₄ is (E°(Cu²⁺|Cu) = 0.34 V; take 2.303RT/F = 0.059 V)
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Answer: (d) 0.281 V

E = E° − (0.059/2) log(1/[Cu²⁺]) = 0.34 − 0.0295 × log 100 = 0.34 − 0.059 = 0.281 V. Adding the term gives 0.399 V; using log 10 gives 0.311 V; dropping the 2 gives 0.222 V.

Question 10
Two hydrogen electrodes, each with H₂ at 1 bar, dip in buffer solutions of pH 2 and pH 5 and are joined by a salt bridge. At 298 K (2.303RT/F = 0.059 V) the emf of the cell is
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Answer: (b) 0.177 V

Each electrode has E = −0.059 pH. The pH 5 electrode (lower E) is the anode: Ecell = (−0.059 × 2) − (−0.059 × 5) = 0.059 × 3 = 0.177 V. 0.413 V adds the two pH values; 0.295 V uses pH 5 alone.

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