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Electrochemistry flash cards

Master Electrochemistry through 86 NEET-level recall cards, systematically structured one idea at a time. Revise concept-wise, identify the areas where you need improvement, and focus your preparation with greater precision.

Electrochemistry, question and answer

22 of this chapter's 86 cards, laid out open so you can read straight through. The remaining 64 are in the interactive deck, where the answer stays hidden until you commit to one.

  1. 1.What is electrochemistry?

    The branch of chemistry that studies the relationship between electrical energy and chemical change — how spontaneous redox reactions produce electricity (galvanic cells) and how electricity drives non-spontaneous reactions (electrolytic cells).

    Hint: Electricity ↔ chemical change

  2. 2.Define oxidation and reduction in terms of electrons.

    Oxidation is loss of electrons (increase in oxidation number); reduction is gain of electrons (decrease in oxidation number). Remember OIL RIG: Oxidation Is Loss, Reduction Is Gain.

    Hint: OIL RIG

  3. 3.What is an electrode potential?

    The tendency of an electrode (metal–ion) to lose or gain electrons when in contact with its ion solution, measured as a potential difference. It arises from the equilibrium Mn++neMM^{n+} + ne^- \rightleftharpoons M.

    Hint: Tendency to be reduced/oxidised

  4. 4.What is a galvanic (voltaic) cell?

    An electrochemical cell that converts the chemical energy of a spontaneous redox reaction into electrical energy. Example: the Daniell cell using ZnZn and CuCu.

    Hint: Spontaneous → electricity

  5. 5.In a galvanic cell, which electrode is the anode and which is the cathode, and their signs?

    Anode = oxidation, negative terminal. Cathode = reduction, positive terminal. Electrons flow from anode to cathode through the external wire.

    Hint: Anode negative in galvanic cell

  6. 6.Describe the Daniell cell and its reactions.

    Zn electrode in ZnSO4ZnSO_4 and Cu electrode in CuSO4CuSO_4. Anode: ZnZn2++2eZn \rightarrow Zn^{2+} + 2e^-. Cathode: Cu2++2eCuCu^{2+} + 2e^- \rightarrow Cu. Overall: Zn+Cu2+Zn2++CuZn + Cu^{2+} \rightarrow Zn^{2+} + Cu, Ecell=1.1 VE^\circ_{cell}=1.1\ V.

    Hint: Zn–Cu, 1.1 V

  7. 7.What is the function of a salt bridge?

    It completes the circuit, allows ion flow between the two half-cells, and maintains electrical neutrality by supplying counter-ions, preventing charge build-up. It also minimises the liquid-junction potential.

    Hint: Neutrality + circuit

  8. 8.Write the cell notation convention (IUPAC).

    Anode written on the left, cathode on the right: AnodeAnode solnCathode solnCathode\text{Anode} \mid \text{Anode soln} \parallel \text{Cathode soln} \mid \text{Cathode}. A single vertical line \mid = phase boundary; double line \parallel = salt bridge.

    Hint: Anode left, cathode right

  9. 9.Write the cell notation for the Daniell cell.

    Zn(s)Zn2+(aq)Cu2+(aq)Cu(s)Zn(s) \mid Zn^{2+}(aq) \parallel Cu^{2+}(aq) \mid Cu(s). Left side is oxidation (anode), right side is reduction (cathode).

    Hint: Zn | Zn2+ || Cu2+ | Cu

  10. 10.What is EMF (electromotive force) of a cell?

    The potential difference between the two electrodes when no current is drawn (open circuit). It is the maximum voltage the cell can deliver.

    Hint: Cell voltage at zero current

  11. 11.Give the formula for standard cell potential EcellE^\circ_{cell}.

    Ecell=EcathodeEanodeE^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}, using standard reduction potentials. Equivalently Ecell=ErightEleftE^\circ_{cell}=E^\circ_{right}-E^\circ_{left}.

    Hint: Cathode minus anode (reduction potentials)

  12. 12.What sign of EcellE^\circ_{cell} indicates a spontaneous cell reaction?

    A positive EcellE^\circ_{cell} means the reaction is spontaneous (galvanic). A negative value means the reaction is non-spontaneous as written.

    Hint: Positive = spontaneous

  13. 13.What is the Standard Hydrogen Electrode (SHE) and its potential?

    The reference electrode PtH2(1 bar)H+(1 M)Pt \mid H_2(1\ bar) \mid H^+(1\ M) with EE^\circ defined as exactly 0.00 V0.00\ V at all temperatures. All other potentials are measured relative to it.

    Hint: Reference, 0.00 V

  14. 14.What is the electrochemical series?

    An arrangement of elements/electrodes in order of their standard reduction potentials EE^\circ. Fluorine (highest positive) is at the top; lithium (most negative) at the bottom.

    Hint: Ranked by E° reduction

  15. 15.What does a more positive EE^\circ indicate?

    A stronger oxidising agent — the species is more easily reduced. F2F_2 (+2.87 V+2.87\ V) is the strongest common oxidising agent; FF^- is a very weak reducing agent.

    Hint: More positive = stronger oxidant

  16. 16.Why can zinc displace copper from CuSO4CuSO_4 but copper cannot displace zinc from ZnSO4ZnSO_4?

    EZn2+/Zn=0.76 VE^\circ_{Zn^{2+}/Zn}=-0.76\ V is more negative than ECu2+/Cu=+0.34 VE^\circ_{Cu^{2+}/Cu}=+0.34\ V, so Zn is the stronger reducing agent and spontaneously reduces Cu2+Cu^{2+}. The reverse is non-spontaneous.

    Hint: Compare E° values

  17. 17.State the Nernst equation for a general electrode Mn++neMM^{n+} + ne^- \rightarrow M.

    E=ERTnFln1[Mn+]E = E^\circ - \dfrac{RT}{nF}\ln\dfrac{1}{[M^{n+}]}, i.e. E=E+RTnFln[Mn+]E = E^\circ + \dfrac{RT}{nF}\ln[M^{n+}] where RR is gas constant, FF Faraday, nn electrons.

    Hint: E = E° − (RT/nF) ln Q

  18. 18.Write the Nernst equation for a full cell reaction with reaction quotient QQ.

    Ecell=EcellRTnFlnQE_{cell} = E^\circ_{cell} - \dfrac{RT}{nF}\ln Q. At 298 K298\ K this becomes Ecell=Ecell0.0591nlogQE_{cell} = E^\circ_{cell} - \dfrac{0.0591}{n}\log Q.

    Hint: −0.0591/n log Q at 298 K

  19. 19.How does the constant 2.303RTF\dfrac{2.303\,RT}{F} evaluate at 298 K298\ K?

    2.303×8.314×298964850.0591 V\dfrac{2.303 \times 8.314 \times 298}{96485} \approx 0.0591\ V. This gives the familiar factor 0.0591/n0.0591/n in the log form of the Nernst equation.

    Hint: ≈ 0.0591 V

  20. 20.For the Daniell cell, write the Nernst equation.

    Ecell=Ecell0.05912log[Zn2+][Cu2+]E_{cell} = E^\circ_{cell} - \dfrac{0.0591}{2}\log\dfrac{[Zn^{2+}]}{[Cu^{2+}]}, since n=2n=2 and Q=[Zn2+]/[Cu2+]Q=[Zn^{2+}]/[Cu^{2+}] (solids omitted).

    Hint: n=2, Q = [Zn2+]/[Cu2+]

  21. 21.How is EcellE_{cell} related to equilibrium constant KK at equilibrium?

    At equilibrium Ecell=0E_{cell}=0 and Q=KQ=K, giving Ecell=0.0591nlogKE^\circ_{cell} = \dfrac{0.0591}{n}\log K at 298 K298\ K, i.e. logK=nEcell0.0591\log K = \dfrac{nE^\circ_{cell}}{0.0591}.

    Hint: E° = (0.0591/n) log K

  22. 22.State the relation between Gibbs energy and cell potential.

    ΔG=nFEcell\Delta G = -nFE_{cell} and ΔG=nFEcell\Delta G^\circ = -nFE^\circ_{cell}, where nn = moles of electrons and F=96485 Cmol1F=96485\ C\,mol^{-1}. Spontaneity (ΔG<0\Delta G<0) corresponds to Ecell>0E_{cell}>0.

    Hint: ΔG = −nFE

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