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B.Sc Chemistry Electrochemical Series — practice questions

Explore 54 free Electrochemical Series questions in B.Sc Chemistry Chemistry. Open each question or exercise to read its content, reference answers and available explanation.

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Electrochemical Series questions and exercises with answers

Showing questions 1–30 of 54.

For a galvanic cell, the standard cell EMF is given by:Read the answer and worked solution →The Nernst equation for a half-cell reaction at 298 K is:Read the answer and worked solution →A device converting CHEMICAL energy to electrical energy via spontaneous redox is a:Read the answer and worked solution →In the Daniell cell, the cathode reaction is:Read the answer and worked solution →EMF of standard Daniell cell at 25°C is approximately:Read the answer and worked solution →Faraday's first law of electrolysis states mass deposited is proportional to:Read the answer and worked solution →The value of Faraday's constant F is approximately:Read the answer and worked solution →In a galvanic cell, electrons flow externally from:Read the answer and worked solution →The Nernst equation gives cell EMF at non-standard conditions. For a half-cell Mⁿ⁺ + ne⁻ → M, it is:Read the answer and worked solution →Relation between standard cell potential E°, ΔG°, and equilibrium constant K:Read the answer and worked solution →For ΔG° < 0 (spontaneous), E° must be:Read the answer and worked solution →The molar conductivity Λ_m of an electrolyte:Read the answer and worked solution →Kohlrausch's law states the limiting molar conductivity of an electrolyte is:Read the answer and worked solution →For the cell Zn | Zn²⁺ (1 M) || Cu²⁺ (1 M) | Cu, with E°(Zn) = -0.76 V, E°(Cu) = +0.34 V, E_cell:Read the answer and worked solution →Standard electrode potential of SHE (Standard Hydrogen Electrode) is taken as:Read the answer and worked solution →For Cu²⁺ + 2e⁻ → Cu, how much charge to deposit 0.5 mol Cu? (F = 96500 C/mol)Read the answer and worked solution →Conductivity κ of pure water is approximately:Read the answer and worked solution →At 25°C, ratio of Nernst factor 0.0591/n to RT/(nF) ln 10 in SI units:Read the answer and worked solution →A current of 2 A flows for 30 min through a CuSO4 solution. Mass of Cu deposited (M_Cu = 63.5):Read the answer and worked solution →For the cell: Zn|Zn²⁺(0.1 M)||Cu²⁺(0.01 M)|Cu, at 25°C, E_cell is:Read the answer and worked solution →Equilibrium constant K of the Daniell cell at 25°C (E° = 1.10 V, n = 2):Read the answer and worked solution →A 0.1 M weak acid HA has Λ_m = 10 S·cm²/mol; at infinite dilution Λ_m^0 = 400. Degree of dissociation α:Read the answer and worked solution →In electrolysis of CuSO4 with Pt electrodes, the cathode product is:Read the answer and worked solution →Standard electrode potentials: E°(Fe³⁺/Fe²⁺) = +0.77 V, E°(I2/I⁻) = +0.54 V. For the reaction 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I2, E° equals:Read the answer and worked solution →Why is the cell potential temperature-dependent?Read the answer and worked solution →For Hg²⁺/Hg (E° = +0.79 V) and Cu²⁺/Cu (E° = +0.34 V), the spontaneous reaction is:Read the answer and worked solution →Conductivity of a 0.01 M NaCl solution is 0.0014 S/cm. Molar conductivity Λ_m:Read the answer and worked solution →Lead-acid battery during DISCHARGE: PbO2 + Pb + 2H2SO4 → 2PbSO4 + 2H2O. The role of PbO2 is:Read the answer and worked solution →Corrosion of iron involves which type of reaction?Read the answer and worked solution →In electroplating silver onto a spoon, the cathode is:Read the answer and worked solution →

Worked practice questions

For a galvanic cell, the standard cell EMF is given by:

  1. $E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} + E^{\circ}_{\text{anode}}$
  2. $E^{\circ}_{\text{cell}} = E^{\circ}_{\text{anode}} - E^{\circ}_{\text{cathode}}$
  3. $E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}$
  4. $E^{\circ}_{\text{cell}} = -2 E^{\circ}_{\text{anode}}$
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Correct answer: C. $E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}$

$E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}$, where both potentials are listed as standard *reduction* potentials. For a spontaneous galvanic cell, $E^{\circ}_{\text{cell}} > 0$, which means the cathode has the higher reduction potential. This is just an algebraic restatement of "reduction happens at the cathode".

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The Nernst equation for a half-cell reaction at 298 K is:

  1. $E = E^{\circ} - n \cdot 0.059 \log Q$
  2. $E = E^{\circ} + \dfrac{0.059}{n} \log Q$
  3. $E = E^{\circ} - \dfrac{0.059}{n} \log Q$
  4. $E = E^{\circ} - 0.059 \log [\mathrm{H^+}]$
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Correct answer: C. $E = E^{\circ} - \dfrac{0.059}{n} \log Q$

Full form: $E = E^{\circ} - \dfrac{RT}{nF} \ln Q$. At 298 K with $\log_{10}$, the constants collapse to $\dfrac{0.0592}{n}$, often rounded to $\dfrac{0.059}{n}$. Here $n$ is the number of electrons exchanged in the half-cell, and $Q$ is the reaction quotient. The negative sign matters: when $Q > 1$ (products dominate), $E$ is below $E^{\circ}$. Option B has the wrong sign; option D scrambles where $n$ goes.

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A device converting CHEMICAL energy to electrical energy via spontaneous redox is a:

  1. Heat engine
  2. Battery charger
  3. Galvanic (voltaic) cell
  4. Electrolytic cell
Show answer and solution

Correct answer: C. Galvanic (voltaic) cell

Galvanic cell (also called voltaic cell, e.g. Daniell cell): spontaneous chemical reaction generates electrical current. **Electrolytic cell does the reverse:** electrical energy drives a non-spontaneous reaction.

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