Answer: a. One Faraday = 96500 C = 1 mol e− = 6.022×10^23 electrons. 9650 C = 9650/96500 = 0.1 mol e− → number = 0.1 × 6.022×10^23 = 6.022×10^22 electrons.
Combine: reduction (cathode) Mn2+ +2e− → Mn (E° = −1.18 V) and oxidation (anode) Mn2+ → Mn3+ + e− (reverse of Mn3+ + e− → Mn2+, so E°ox = +1.51 V). E°cell = E°cathode − E°anode = (−1.18) − (−1.51) = +0.33 V. Positive → spontaneous.
E°cell = E°cathode − E°anode = 0.34 − (−0.76) = 1.10 V.
Λm (S cm^2 mol−1) = (κ × 1000) / c(mol dm−3). So Λm = (5.76×10−3 ×1000)/0.5 = 11.52 S cm^2 mol−1.
Use additivity: Λ°(HOAc) = Λ°(HCl) + Λ°(NaOAc) − Λ°(NaCl) = 426.2 + 91.0 − 126.5 = 390.7 S cm^2 mol−1.
Faraday (F) is the charge carried by one mole of electrons (≈ 96500 C mol−1). (Option d is equivalent numerically, but definition best given by b.)
Mn changes from +7 to +2, a gain of 5 electrons per Mn; thus 5 faradays are required per mole of MnO4− reduced to Mn2+.
t = 41×60+40 = 2500 s. Q = I t = 3.86×2500 = 9650 C = 0.1 F. Ca2+ requires 2 F per mole Ca. moles Ca = Q/(2F)=0.1/2=0.05 mol. Mass = 0.05×40 = 2 g.
To form 0.1 mol Cl2, electrons required = 2×0.1 = 0.2 mol e− → Q = 0.2×96500 = 19300 C. t = Q/I = 19300/3 = 6433.3 s = 107.22 min.
Q = I t = 1×60 = 60 C. Number of electrons = Q / e = 60 / (1.6×10−19) ≈ 3.75×10^20 electrons.
Specific conductance is proportional to concentration; the smallest concentration (0.002 N) gives the least conductance.
Answer: a. During charging the negative plate (cathode) is reduced: PbSO4 + 2e− → Pb + SO4^2−. The positive plate is oxidised to PbO2. Option (a) correctly describes the cathode change.
Leclanché and mercury cells are primary (non‑rechargeable); Ni–Cd and lead storage are secondary (rechargeable).
Zn2+/Zn has a more negative standard potential (≈ −0.76 V) than Fe2+/Fe (≈ −0.44 V). Zinc is more easily oxidized and acts as sacrificial anode protecting iron; hence Zn coats Fe, not vice versa.
Correct facts: heating iron in dry air gives a black oxide layer mainly Fe3O4 (not the hydrated red‑brown rust). Rust (the flaky hydrated iron(III) oxide) is generally Fe2O3·nH2O. The assertion as stated (formation of 'rust' on heating in dry air) is not correct; the reason statement as given here (if interpreted as 'rust = Fe2O3' rather than hydrated oxide) is an oversimplification. (Given ambiguity in the printed choices, answer selected with low confidence.)
Answer: a. In alkaline fuel cells oxygen is reduced at the cathode: O2 + 2H2O + 4e− → 4OH−.
Degree of dissociation α = Λ/Λ∞ = 6/400 = 0.015. For a weak monobasic acid Ka ≈ cα^2 = (3.0×10−3)(0.015)^2 ≈ 6.75×10−7. Nearest option: 6.25×10−7.
Cell constant K = κ × R = 1.25×10−3 S cm−1 × 800 Ω = 1.0 cm−1.
Method: convert Λ∞ to SI units (S m^2 mol−1), find s = κ/Λ∞, then Ksp = s^2. Using typical Λ∞ ≈126.5 S cm^2 mol−1 (→0.01265 S m^2 mol−1) gives s ≈ κ/Λ∞ and Ksp ≈ (κ/Λ∞)^2 ≈ 1.7×10−6 (approximately). Selected nearest option. (Details depend on the exact Λ∞ used; answer given with low confidence.)
Nernst: E = E° − (0.059/2) log([Zn2+]/[Cu2+]). For case 1: log(0.01/1) = −2 → E1 = E° +0.059. For case 2: log(1/0.01)=+2 → E2 = E° −0.059. Hence E1 > E2.
A species disproportionates when its standard potential lies between the potentials for its oxidation and reduction so it can both oxidize and reduce itself. Hypobromous acid (HOBr / HBrO) is known to disproportionate to bromide and bromate (e.g. 3HOBr → 2Br- + BrO3- + 3H+). Thus HBrO undergoes disproportionation.
ΔG° = -nFE°. For the given cell n = 2 (two electrons transferred). Using F = 96485 C mol-1, ΔG° = -2 × 96485 × 0.24 = -46312 J mol-1 ≈ -46.312 kJ mol-1 ≈ -46.32 kJ mol-1.
Mass H2 = 0.504 g → n(H2) = 0.504/2 = 0.252 mol. Electrons used: 2 e- per H2 → n(e-) = 0.252×2 = 0.504 mol e-. For Cu2+ + 2e- → Cu, moles Cu deposited = 0.504/2 = 0.252 mol. Mass Cu ≈ 0.252 × 63.5 ≈ 16.0 g (≈ 15.8 g given as closest choice).
A species with a higher reduction potential will oxidize a species with lower reduction potential. Since E°(Y) > E°(X), Y (as the oxidant) can oxidize X. But E°(Z) > E°(Y), so Z will oxidize Y. Thus Y will oxidize X and not Z.
Overall reaction transfers n = 2 e-. From log K: E°cell = (2.303RT/nF)·log K. Using R=8.314 J mol-1K-1, T=300 K, F=96485 C mol-1, factor = 2.303·8.314·300/96485 ≈ 0.05956 V. So E°cell = 0.05956×15.6/2 ≈ 0.4646 V. For the cell, E°cell = E°(cathode) - E°(anode) = E°(B+/B) - E°(A2+/A). Hence E°(B+/B) = E°cell + E°(A2+/A) = 0.4646 + 0.34 ≈ 0.8046 V ≈ 0.80 V.
Anode: electrode where oxidation occurs (electrons are produced). Cathode: electrode where reduction occurs (electrons are consumed). In a galvanic cell the anode is negative and the cathode is positive; in an electrolytic cell polarities are reversed.
Anode: electrode where oxidation occurs (electrons are produced). Cathode: electrode where reduction occurs (electrons are consumed). In a galvanic cell the anode is negative and the cathode is positive; in an electrolytic cell polarities are reversed.
Specific conductance κ depends on ionic concentration and mobility. On dilution the number of charge carriers per unit volume decreases, so κ decreases (even though molar conductivity increases due to reduced inter-ionic interactions).
Specific conductance κ depends on ionic concentration and mobility. On dilution the number of charge carriers per unit volume decreases, so κ decreases (even though molar conductivity increases due to reduced inter-ionic interactions).
Kohlrausch's law of independent migration: at infinite dilution the molar conductivity Λm° of an electrolyte equals the sum of the ionic contributions: Λm° = ν+ λ+° + ν- λ-°, where λ±° are limiting ionic molar conductivities and ν± stoichiometric numbers. For a weak electrolyte, Λm measured at various dilutions can be extrapolated to infinite dilution, or Λm° obtained by adding λ° values of constituent ions determined from strong electrolytes; then degree of dissociation and K can be evaluated.
Kohlrausch's law of independent migration: at infinite dilution the molar conductivity Λm° of an electrolyte equals the sum of the ionic contributions: Λm° = ν+ λ+° + ν- λ-°, where λ±° are limiting ionic molar conductivities and ν± stoichiometric numbers. For a weak electrolyte, Λm measured at various dilutions can be extrapolated to infinite dilution, or Λm° obtained by adding λ° values of constituent ions determined from strong electrolytes; then degree of dissociation and K can be evaluated.
Molten NaCl contains Na+ and Cl-. At cathode (reduction): Na+ + e- → Na(l). At anode (oxidation): 2Cl- → Cl2(g) + 2e-. Products: liquid sodium at cathode and chlorine gas at anode. Electrodes are inert (e.g. graphite) to avoid side reactions.
Molten NaCl contains Na+ and Cl-. At cathode (reduction): Na+ + e- → Na(l). At anode (oxidation): 2Cl- → Cl2(g) + 2e-. Products: liquid sodium at cathode and chlorine gas at anode. Electrodes are inert (e.g. graphite) to avoid side reactions.
Faraday's first law: mass of substance liberated at an electrode is directly proportional to the total charge passed, m ∝ Q. Second law: for the same quantity of electricity, masses of substances liberated are proportional to their chemical equivalent weights. Quantitatively: m = (Q·M)/(n·F) where M = molar mass, n = electrons per formula, F = Faraday constant.
Faraday's first law: mass of substance liberated at an electrode is directly proportional to the total charge passed, m ∝ Q. Second law: for the same quantity of electricity, masses of substances liberated are proportional to their chemical equivalent weights. Quantitatively: m = (Q·M)/(n·F) where M = molar mass, n = electrons per formula, F = Faraday constant.
Daniel cell: two half-cells: Zn(s) | Zn2+(aq) (usually ZnSO4) and Cu(s) | Cu2+(aq) (CuSO4). Salt bridge (porous pot or KNO3 bridge) completes circuit. Anode (Zn): Zn → Zn2+ + 2e- (oxidation). Cathode (Cu): Cu2+ + 2e- → Cu (reduction). Overall: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s).
Daniel cell: two half-cells: Zn(s) | Zn2+(aq) (usually ZnSO4) and Cu(s) | Cu2+(aq) (CuSO4). Salt bridge (porous pot or KNO3 bridge) completes circuit. Anode (Zn): Zn → Zn2+ + 2e- (oxidation). Cathode (Cu): Cu2+ + 2e- → Cu (reduction). Overall: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s).
In a galvanic cell oxidation at the anode releases electrons into the electrode, making it electron-rich (negative). Electrons flow through the external circuit to the cathode where reduction consumes electrons, making the cathode electron-deficient relative to anode (positive).
In a galvanic cell oxidation at the anode releases electrons into the electrode, making it electron-rich (negative). Electrons flow through the external circuit to the cathode where reduction consumes electrons, making the cathode electron-deficient relative to anode (positive).
(i) Λm = (κ×1000)/c = (1.5×10^-4 ×1000)/0.01 = 15.0 S cm^2 mol^-1. (ii) Degree of dissociation α = Λm/Λm° = 15.0/248.2 = 0.0604. Dissociation constant K = c α^2/(1−α) = 0.01×(0.0604)^2/(0.9396) ≈ 3.9×10^-5. (Values rounded to two significant figures.)
(i) Λm = (κ×1000)/c = (1.5×10^-4 ×1000)/0.01 = 15.0 S cm^2 mol^-1. (ii) Degree of dissociation α = Λm/Λm° = 15.0/248.2 = 0.0604. Dissociation constant K = c α^2/(1−α) = 0.01×(0.0604)^2/(0.9396) ≈ 3.9×10^-5. (Values rounded to two significant figures.)
0.1 M HCl has greater molar conductivity. Although both are strong electrolytes at same concentration, H+ has very high ionic mobility (very large λ°) compared to K+, so Λm(HCl) > Λm(KCl).
0.1 M HCl has greater molar conductivity. Although both are strong electrolytes at same concentration, H+ has very high ionic mobility (very large λ°) compared to K+, so Λm(HCl) > Λm(KCl).
Specific conductance κ generally increases with concentration (up to moderate concentrations). Decreasing order: v (0.5 M) > iv (0.25 M) > iii (0.1 M) > i (0.01 M) > ii (0.005 M).
Specific conductance κ generally increases with concentration (up to moderate concentrations). Decreasing order: v (0.5 M) > iv (0.25 M) > iii (0.1 M) > i (0.01 M) > ii (0.005 M).
DC causes electrode polarization and electrolysis (chemical changes at electrodes) altering concentrations and producing gases. AC (usually at suitable frequency) avoids net electrochemical reaction and polarization, giving stable conductance measurements.
DC causes electrode polarization and electrolysis (chemical changes at electrodes) altering concentrations and producing gases. AC (usually at suitable frequency) avoids net electrochemical reaction and polarization, giving stable conductance measurements.
Specific conductance κ is a property of the solution and is independent of the cell constant. Both cells measure the same κ (assuming ideal electrodes and identical solution and temperature); cell constant affects measured conductance G but κ = G × (cell constant) yields the same κ.
Specific conductance κ is a property of the solution and is independent of the cell constant. Both cells measure the same κ (assuming ideal electrodes and identical solution and temperature); cell constant affects measured conductance G but κ = G × (cell constant) yields the same κ.
Initial moles Cu2+ = 0.250 L × 0.5 M = 0.125 mol. Charge passed Q = I·t = 1.608 A × (50 min × 60 s/min = 3000 s) = 4824 C. Moles e- = Q/F = 4824/96485 ≈ 0.0500 mol e-. Moles Cu2+ reduced = (moles e-)/2 = 0.0250 mol. Remaining Cu2+ = 0.125 − 0.025 = 0.100 mol. Concentration = 0.100 mol / 0.250 L = 0.400 M.
Initial moles Cu2+ = 0.250 L × 0.5 M = 0.125 mol. Charge passed Q = I·t = 1.608 A × (50 min × 60 s/min = 3000 s) = 4824 C. Moles e- = Q/F = 4824/96485 ≈ 0.0500 mol e-. Moles Cu2+ reduced = (moles e-)/2 = 0.0250 mol. Remaining Cu2+ = 0.125 − 0.025 = 0.100 mol. Concentration = 0.100 mol / 0.250 L = 0.400 M.
Consider Fe3+ + Br- → Fe2+ + 1/2 Br2. E°cell = E°(reduction at cathode) − E°(reduction at anode) = E°(Fe3+/Fe2+) − E°(Br2/Br-) = 0.771 − 1.09 = −0.319 V (negative). Thus the reaction is non-spontaneous under standard conditions. Fe3+ cannot oxidize Br- to Br2 under standard conditions.
Consider Fe3+ + Br- → Fe2+ + 1/2 Br2. E°cell = E°(reduction at cathode) − E°(reduction at anode) = E°(Fe3+/Fe2+) − E°(Br2/Br-) = 0.771 − 1.09 = −0.319 V (negative). Thus the reaction is non-spontaneous under standard conditions. Fe3+ cannot oxidize Br- to Br2 under standard conditions.
Reaction: Cu2+ + Fe → Cu + Fe2+. E°cell = E°(Cu2+/Cu) − E°(Fe2+/Fe) = 0.34 − (−0.44) = +0.78 V (positive), so spontaneous. Fe metal will be oxidized and Cu will plate out. Therefore copper sulphate cannot be safely stored long-term in an iron vessel.
Reaction: Cu2+ + Fe → Cu + Fe2+. E°cell = E°(Cu2+/Cu) − E°(Fe2+/Fe) = 0.34 − (−0.44) = +0.78 V (positive), so spontaneous. Fe metal will be oxidized and Cu will plate out. Therefore copper sulphate cannot be safely stored long-term in an iron vessel.
A metal will displace hydrogen from acid if it is more easily oxidised than H (i.e. its reduction potential is more negative than 0.00 V). Since E°(M1) = -x V < 0, M1 is more reactive than H and will oxidise to M1n+ releasing H2. M2 with positive E° is less reactive than H and will not liberate H2 from H2SO4.
M1 (the metal with negative reduction potential) will liberate H2 from dilute H2SO4; M2 (with positive E°) will not.
For electroplating onto iron the coating metal must be reduced preferentially on Fe (not allow Fe to oxidise). That requires E°(coating metal) > E°(Fe2+/Fe). Therefore whichever of M1 or M2 has E° > −0.44 V will plate on iron and serve as a coating. (If E°(M) < −0.44 V, iron will be oxidised instead and coating won't form.)
The metal whose E° is more positive than −0.44 V (i.e. more noble than Fe) is better for coating iron.
Cell notation: Cd(s)|Cd2+||Cu2+|Cu(s). Cathode is the half-cell with higher (more positive) E°: Cu2+/Cu (0.34 V). Anode: Cd2+/Cd (−0.40 V). E°cell = E°cathode − E°anode = 0.34 − (−0.40) = 0.74 V. Oxidation at anode: Cd → Cd2+ + 2e−. Reduction at cathode: Cu2+ + 2e− → Cu. Net: Cd(s) + Cu2+ → Cd2+ + Cu(s). Since E°cell > 0, the reaction is spontaneous.
E°cell = 0.74 V. Cell reaction: Cd(s) + Cu2+ → Cd2+ + Cu(s). Reaction is spontaneous (feasible).
Molar volume at 25°C ≈ 24.45 L mol−1. Moles H2 = 44.8 / 24.45 = 1.832 mol. Each mole H2 → 2 e−, so total charge Q = n( H2) × 2F = 1.832 × 2 × 96485 = 3.536×10^5 C. Time = 10 min = 600 s. Current I = Q/t = 3.536×10^5 / 600 ≈ 589 A. For Cu2+ + 2e− → Cu, moles Cu deposited = moles H2 (since 2 e− per H2 and 2 e− per Cu) = 1.832 mol. Mass = 1.832 × 63.55 ≈ 116.5 g.
Average current ≈ 589 A. Mass of Cu deposited ≈ 116.5 g.
Moles Ni deposited = 2.935 / 58.74 = 0.05000 mol. Ni2+ + 2e− → Ni so electrons used = 0.05000 × 2 = 0.10000 mol e−. Same total electrons deposit Cr3+ via Cr3+ + 3e− → Cr, so moles Cr = 0.10000 / 3 = 0.033333 mol. Mass Cr = 0.033333 × 52.0 = 1.733 g.
Mass of Cr deposited = 1.733 g (approx).
Nernst equation for Cu2+ + 2e− → Cu: E = E° + (0.0591/2) log[Cu2+]. Substitute: E = 0.34 + (0.0591/2) log(0.10) = 0.34 + 0.02955 × (−1) = 0.340 − 0.02955 = 0.31045 V ≈ 0.310 V.
E = 0.310 V (approximately).
E°cell = E°(cathode) − E°(anode) = 0.80 − (−2.37) = 3.17 V. Net reaction: Mg + 2Ag+ → Mg2+ + 2Ag, so n = 2. At 25°C, E° = (0.0591/n) log K ⇒ log K = nE°/0.0591 = 2×3.17/0.0591 ≈ 107.3 ⇒ K ≈ 10^107.3 (extremely large, reaction essentially complete). Maximum electrical work per mole reaction: w_max = −nFE° = −2 × 96485 C mol−1 × 3.17 V ≈ −6.11×10^5 J mol−1.
E°cell = 3.17 V. Equilibrium constant K ≈ 10^107.3 (very large). Maximum work w_max = −nFE° ≈ −6.11×10^5 J per mole of reaction (n=2).
Moles of water = (9.2×10^12 L) × (1000 g L−1) / 18 g mol−1 = 9.2×10^15 / 18 ≈ 5.111×10^14 mol. Electrolysis: 2 H2O → 2 H2 + O2 requires 4 e− per 2 H2O ⇒ 2 e− per H2O. Charge needed Q = (moles H2O) × 2F = 5.111×10^14 × 2 × 96485 ≈ 9.87×10^19 C. At I = 2.0×10^6 C s−1, time t = Q/I = 9.87×10^19 / 2.0×10^6 = 4.935×10^13 s = 4.935×10^13 / (3.1536×10^7 s yr−1) ≈ 1.57×10^6 years. Note: numeric result depends on the assumed current; replace I if different.
Approximately 1.6×10^6 years (about 1.57 million years) under the stated assumption I = 2.0×10^6 C s−1.
Start from thermodynamics: ΔG = ΔG° + RT ln Q. For an electrochemical cell, ΔG = −nFE and ΔG° = −nFE°. Substitute: −nFE = −nFE° + RT ln Q ⇒ divide by −nF ⇒ E = E° − (RT/nF) ln Q. Converting ln to log10 and inserting numerical constant at 298.15 K gives E = E° − (2.303RT/nF) log10 Q = E° − (0.05916/n) log10 Q (≈0.0591/n at 25°C).
E = E° − (RT/nF) ln Q (general form); at 25°C E = E° − (0.0591/n) log10 Q.
In sacrificial (cathodic) protection a metal with a more negative electrode potential (e.g. Zn, Mg, or Al) is electrically connected to iron or steel. The active metal serves as the anode and corrodes (is 'sacrificed'), while the iron becomes the cathode and is protected from oxidation. Applications: galvanising (Zn coating on iron), sacrificial anodes on ships, underground pipelines and storage tanks. Advantages: simple and effective; requires periodic replacement of sacrificial anode.
Sacrificial protection uses a more active metal (anodic) attached to iron/steel so the active metal corrodes preferentially and protects the iron.
Typical proton-exchange membrane (PEM) fuel cell reactions: Anode: H2 → 2H+ + 2e−. Cathode: 1/2 O2 + 2H+ + 2e− → H2O. Net: H2 + 1/2 O2 → H2O; E°cell ≈ 1.23 V (standard). Electrons flow through the external circuit producing current; protons migrate through the membrane. Catalysts (Pt) enhance kinetics. Products are electricity and water; fuel cells have high efficiency and low pollution (no CO2 if H2 is from non-carbon source).
H2–O2 fuel cell converts chemical energy of H2 + 1/2 O2 → H2O into electrical energy via separate oxidation (anode) and reduction (cathode) half-reactions with an external circuit.
For Al2(SO4)3: ν(Al3+) = 2, ν(SO42−) = 3. Ionic conductances are given per equivalent. Molar ionic conductance = z × (ionic conductance per equiv). For Al3+: z = 3 ⇒ contribution = 2 × 3 × 189 = 1134 mho·cm2·mol−1. For SO42−: z = 2 ⇒ contribution = 3 × 2 × 160 = 960 mho·cm2·mol−1. Molar conductance Λ°m = 1134 + 960 = 2094 mho·cm2·mol−1. Number of equivalents per mole of Al2(SO4)3 = total positive charge = 2×3 = 6. Hence Λ°eq = Λ°m / 6 = 2094 / 6 = 349 mho·cm2·equiv−1.
Equivalent conductance Λ°eq = 349 mho·cm2·equiv−1. Molar conductance Λ°m = 2094 mho·cm2·mol−1.