Level 1 — RecognitionRedox & Electrochemistry (Intro)

Redox & Electrochemistry (Intro)

⏱ 20 minutes30 marksprintable — key stays hidden on paper

Time limit: 20 minutes Total marks: 30

Section A — Multiple Choice (1 mark each)

Choose the single best answer.

Q1. In a galvanic cell, oxidation occurs at the:

(a) cathode
(b) anode
(c) salt bridge
(d) voltmeter

Q2. The standard hydrogen electrode (SHE) is assigned a standard potential of:

(a) $+1.00\text{ V}$
(b) $-0.76\text{ V}$
(c) $0.00\text{ V}$
(d) $+0.34\text{ V}$

Q3. The correct expression for standard cell EMF is:

(a) $E^\circ_{cell} = E^\circ_{anode} - E^\circ_{cathode}$
(b) $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$
(c) $E^\circ_{cell} = E^\circ_{cathode} + E^\circ_{anode}$
(d) $E^\circ_{cell} = -E^\circ_{cathode} - E^\circ_{anode}$

Q4. For a spontaneous cell reaction, the values of $E^\circ_{cell}$ and $\Delta G$ are:

(a) $E^\circ_{cell} > 0$ , $\Delta G > 0$
(b) $E^\circ_{cell} < 0$ , $\Delta G < 0$
(c) $E^\circ_{cell} > 0$ , $\Delta G < 0$
(d) $E^\circ_{cell} = 0$ , $\Delta G = 0$

Q5. The relationship between $\Delta G^\circ$ and $E^\circ_{cell}$ is:

(a) $\Delta G^\circ = nFE^\circ$
(b) $\Delta G^\circ = -nFE^\circ$
(c) $\Delta G^\circ = -nF/E^\circ$
(d) $\Delta G^\circ = RT\ln E^\circ$

Q6. In the Nernst equation $E = E^\circ - \dfrac{RT}{nF}\ln Q$ , at $298\text{ K}$ the factor $\dfrac{RT}{F}$ (converted to base-10) gives approximately:

(a) $0.0592\text{ V}$
(b) $0.592\text{ V}$
(c) $0.0296\text{ V}$
(d) $1.00\text{ V}$

Q7. A concentration cell generates an EMF because:

(a) the two electrodes are made of different metals
(b) $E^\circ_{cell} \neq 0$
(c) the ion concentrations in the two half-cells differ
(d) a fuel is continuously supplied

Q8. Which is a secondary (rechargeable) battery?

(a) Leclanché dry cell
(b) Lead-acid accumulator
(c) $H_2/O_2$ fuel cell
(d) Zinc–carbon cell

Q9. In a $H_2/O_2$ fuel cell, the product formed is:

(a) $CO_2$
(b) $H_2O$
(c) $H_2O_2$
(d) $O_3$

Q10. Faraday's first law of electrolysis is expressed as:

(a) $m = ZIt$
(b) $m = ZI/t$
(c) $m = Z/(It)$
(d) $m = It/Z$

Q11. In the electrolytic extraction of aluminium, alumina is dissolved in molten:

(a) $NaCl$
(b) cryolite
(c) water
(d) $CaCO_3$

Q12. Galvanization protects iron from rusting by coating it with:

(a) copper
(b) tin
(c) zinc
(d) chromium

Section B — Matching (1 mark each row; 4 marks)

Q13. Match each species/electrode in Column X with its role in Column Y.

Column X	Column Y
(i) Anode	(P) reduction half-reaction
(ii) Cathode	(Q) reference electrode, $E^\circ = 0$
(iii) SHE	(R) oxidation half-reaction
(iv) Salt bridge	(S) maintains electrical neutrality

Section C — True/False with Justification (2 marks each: 1 verdict + 1 reason; 14 marks)

Q14. Electrons flow through the external circuit from anode to cathode in a galvanic cell.

Q15. A metal higher (more negative $E^\circ$ ) in the electrochemical series is a weaker reducing agent than one lower down.

Q16. For the reaction with $E^\circ_{cell} = +1.10\text{ V}$ and $n = 2$ , the equilibrium constant $K$ is greater than 1.

Q17. In a lead-acid battery during discharge, the lead electrode acts as the cathode.

Q18. In a concentration cell, oxidation occurs at the electrode dipping in the more dilute solution.

Q19. Passing the same quantity of charge through solutions of $Ag^+$ and $Cu^{2+}$ deposits equal numbers of moles of each metal.

Q20. Attaching a block of magnesium to an underground steel pipeline provides cathodic protection.

Answer keyMark scheme & solutions

Section A (12 marks)

Q1 — (b) anode. Oxidation (loss of electrons) occurs at the anode by definition; mnemonic "an-ox". (1)

Q2 — (c) $0.00\text{ V}$ . SHE is the arbitrary zero reference for all electrode potentials. (1)

Q3 — (b) $E^\circ_{cathode} - E^\circ_{anode}$ . EMF = reduction potential of cathode minus that of anode. (1)

Q4 — (c) $E^\circ_{cell} > 0$ , $\Delta G < 0$ . Since $\Delta G = -nFE$ , positive EMF gives negative $\Delta G$ = spontaneous. (1)

Q5 — (b) $\Delta G^\circ = -nFE^\circ$ . Standard relation linking free energy and cell potential. (1)

Q6 — (a) $0.0592\text{ V}$ . $\dfrac{RT}{F}\ln 10 = \dfrac{2.303RT}{F} \approx 0.0592\text{ V}$ at 298 K. (1)

Q7 — (c) differing ion concentrations. EMF arises from concentration difference driving the system toward equal concentrations; $E^\circ_{cell}=0$ . (1)

Q8 — (b) Lead-acid accumulator. It is rechargeable (reversible reaction). (1)

Q9 — (b) $H_2O$ . Overall: $2H_2 + O_2 \to 2H_2O$ . (1)

Q10 — (a) $m = ZIt$ . Mass deposited $\propto$ charge $(Q = It)$ . (1)

Q11 — (b) cryolite. $Na_3AlF_6$ lowers the melting point of alumina (Hall–Héroult). (1)

Q12 — (c) zinc. Zinc is more reactive; it corrodes sacrificially (sacrificial protection). (1)

Section B (4 marks)

Q13. (i)→R, (ii)→P, (iii)→Q, (iv)→S. (1 each)

Section C (14 marks)

Q14 — TRUE. (verdict 1) Anode is negative in a galvanic cell; electrons released by oxidation travel externally to the cathode where reduction consumes them. (reason 1)

Q15 — FALSE. (1) More negative $E^\circ$ means the metal loses electrons more readily, so it is a stronger reducing agent, not weaker. (1)

Q16 — TRUE. (1) $\ln K = \dfrac{nFE^\circ}{RT}$ ; since $E^\circ>0$ , $\ln K>0$ , so $K>1$ . Numerically $\log K = \dfrac{2(1.10)}{0.0592}\approx 37$ , so $K \gg 1$ . (1)

Q17 — FALSE. (1) During discharge $Pb$ is oxidized ( $Pb \to Pb^{2+}+2e^-$ ), so it is the anode; $PbO_2$ is the cathode. (1)

Q18 — TRUE. (1) The cell moves toward equal concentration: the dilute side must increase its ion concentration, so the metal there dissolves (oxidation = anode). (1)

Q19 — FALSE. (1) Same charge deposits equal equivalents, not moles. $Ag^+ + e^- \to Ag$ ( $n=1$ ) but $Cu^{2+}+2e^- \to Cu$ ( $n=2$ ), so twice as many moles of Ag as Cu. (1)

Q20 — TRUE. (1) Magnesium (more negative $E^\circ$ ) becomes the sacrificial anode, forcing the steel to be the cathode and preventing its oxidation. (1)

[
  {"claim":"RT/F * ln10 approx 0.0592 V at 298 K (Q6)","code":"R=8.314; T=298; F=96485; val=(R*T/F)*ln(10); result = abs(float(val)-0.0592) < 0.0005"},
  {"claim":"log10 K approx 37 for E=1.10V, n=2 (Q16)","code":"E=Rational(110,100); n=2; logK=n*E/Rational(592,10000); result = abs(float(logK)-37.16) < 1.0"},
  {"claim":"K>1 when E>0 (Q16)","code":"E=1.10; n=2; F=96485; R=8.314; T=298; K=exp(n*F*E/(R*T)); result = float(K) > 1"},
  {"claim":"Ag:Cu mole ratio = 2:1 for equal charge (Q19)","code":"nAg=1; nCu=2; ratio=Rational(nCu,nAg); result = ratio == 2"}
]