Electrochemistry

ELECTRO CHEMISTRY

• Deals with the relationship between current electricity and chemical reactions.
• Electrolytes: Substances that dissociate into ions in solution.
  • Strong electrolytes: NaCl, KCl, Na2SO4, AgNO3, CuSO4, ZnSO4
  • Weak electrolytes: CH3COOH, NH4OH

Electrochemistry

• Electrolytic cell:
  • Chemical reactions occur when current is passed through the solution of an electrolyte.
• Galvanic cell:
  • Current is generated due to the occurrence of certain chemical reactions.

ELECTROLYTIC CELL

• At cathode (Reduction occurs):
  • $Na^+(l) + e^- → Na(s)$
• At anode (Oxidation occurs):
  • $Cl^-(l) → {1 \over 2} Cl_2 (g) + e^-$
• Overall reaction:
  • $Na^+(l) + Cl^- (l) → Na (s) + {1 \over 2} Cl_2 (g)$
• Electrolysis:
  • Movement of ions of electrolyte towards respective electrodes and their subsequent reduction and oxidation when current is passed through electrolytic solution.

Preferential Discharge Theory

• If two cations approach, the cation with a higher reduction potential gets reduced preferentially.

• Reduction Potential: Tendency to get reduced.

  • Order of reduction potential: Na^+ < Mg^{2+} < Al^{3+} < Zn^{2+} < H^+ < Cu^{2+} < Ag^+

• Reduction at Cathode: $Na^+(aq) + e^- →$

• Oxidation at Anode: $H^+(aq) + e^- →, Cl^- (aq) →, OH^- (aq) →$

• Oxidation Potential: Tendency to get oxidized

  • Order of oxidation potential: SO4^{2-} < CO3^{2-} < NO_3^- < Cl^- < OH^-
  • $2OH^- (aq) → H<em>2O(l) + {1 \over 2} O</em>2(g) + 2e^-$

• Aq. CuSO4 solution

  • Cathode: $Cu^{2+} + 2e^- → Cu(s)$
  • Anode: $2H<em>2O(l) → O</em>2(g) + 4H^+ + 4e^-$

• Aq. Na2SO4 solution

  • Cathode: $H_2O →$
  • Anode: $2OH^- → {1 \over 2} H<em>2O (l) + {1 \over 2} O</em>2(g) + 2e^-$

• Aq. AgNO3 solution

  • Cathode: $Ag^+ + e^- → Ag(s)$
  • Anode: $2OH^- → {1 \over 2} H<em>2O + {1 \over 2} O</em>2 +2e^-$

• At Cathode: $Na^+(g) + e^- → Na-Hg$

• At Anode: $Cl^- → {1 \over 2} Cl_2 (g) + e^-$ or $OH^- →$

• Electrode: Hg/ Hg2Cl2 Calomel electrode

• At Cathode (Reduction):

  • $Zn^{2+} (g) + 2e^- → Zn (s)$

• At Anode (Oxidation):

  • $Zn (s) → Zn^{2+} (g) + 2e^-$

• Homework

  • Aq. ZnSO4 solution
    • Cathode: Pt
    • Anode: Hg2Cl2
  • Aq. Na2SO4 solution
    • Cathode:
    • Anode:

• Revise Notes 2 times

• Complete Homework.

• Best of luck

ELECTROLYTIC CONDUCTANCE

• Conductance is the flow of charge.

• Electrolytic resistance arises due to:

  • Solute – solute interaction: Force of attraction between cation & anion of electrolytes.
  • Solute – solvent interaction (Electrophoretic effect)

• Electrolytic conductance arises due to the flow of ions.

• Resistance is the opposition offered in the flow of charge.

• Factors affecting electrolytic Resistance:

  • $R ∝ l$
  • $R ∝ {1 \over A}$
  • $R ∝ {l \over A}$
  • $R = ∫ . {l \over A}$
  • K= Conductivity / specific conductivity
  • C= Conductance
  • ${{l \over A}} = cell constant$
  • ${1 \over ∫} = {1 \over R} . {l \over A}$
  • $𝐾 = 𝐶. {l \over A}$

Physical significance of specific conductivity (K)

• Specific conductivity (K) of an electrolytic solution is equal to conductance offered by unit volume of electrolytic solution when electrodes are unit distance apart.

• $K= 𝐶. {l \over A}$

  • If l = 1cm & A = 1cm2
  • $𝐾 = 𝐶. {1 \over 1}$

• Unit of ‘K’

  • $𝐾 = 𝐶. {l \over A}$
  • $𝐾 = {cm \over cm^2}$
  • $𝐾 = ohm^{-1}cm^{-1}$
  • K (NaCl solution) = 20 Scm-1
  • $𝐾 = Sm^2$
  • $Scm^2 = 100 Sm^2$

• Cal. Specific conductivity of KBr solution having resistance of $100 \Omega$ in an electrolytic cell of length = 10 cm & area of cross section $100 cm^2$.

MOLAR CONDUCTIVITY ($\Lambda_m$)

• It is the total conductance offered by 1 mole of an electrolyte when electrodes are unit distance apart.

• ‘M’ moles of electrolyte → 1000 cc

• 1 moles electrolyte → $\frac{100}{m}$ cc solution

• 1cc solution → K

• ${1000 \over m} cc solution → k {1 \over 1} × {1000 \over m}$

• $\Lambda_𝑚 = 𝐾 × {1000 \over M}$

• Unit of $\Lambda_m$

  • $\Lambda_𝑚 = 𝐾 × {1000 \over M}$
  • $\Lambda_𝑚 = Scm^{-1} {Mol × cm^3}$
  • $\Lambda_𝑚 = Scm^{-1}mol^{-1}$
  • Another unit of$\Lambda_𝑚$
    • $Scm^2mol^{-1}$
    • $1Sm^2 mol^{-1} = 10^4 Scm^2mol^{-1}$

• Cal. $\Lambda_𝑚$ of 0.1 M NaBr solution whose resistance is found to be $200\Omega$ in an electrolytic cell having cell constant 0.1 cm-1 .

EFFECT OF DILUTION ON SPECIFIC CONDUCTIVITY (K)

• On dilution, specific conductivity of an electrolytic solution decreases because the number of ions per unit volume decreases on dilution.

• Cal. ‘K’ of 0.05 M KCl solution whose molar conductivity is 100 $sm^2mol^{-1}$ in a cell having cell constant 0.01 m-1 .

  • A. 10 Scm-1
  • B. 1/100 Scm-1
  • C. 1/1000 Scm-1
  • D. none

• Revise Notes 2 times

• Complete Homework.

• NCERT Reading

• Best of luck

• Electrolytic conductance arises due to flow of ions.

• $K = cl$

• Sem-1/sm-1

• $S cm^2 = 100 sm^{-1}$

• $\Lambda M= K {1000 \over M}$

• $Scm^2 mor^{-1} or Sm^2 moi^{-1}$

• Sm^2 mal ^{-1} => 10^4 Semme

EFFECT OF DILUTION ON MOLAR CONDUCTIVITY

• Debye. Haeckel – Onsanger theory

  • This theory gives the variation of $\Lambda_m$ of strong electrolytes with dilution
  • $\Lambda<em>m NaCl = \lambda</em>m Na^+ + \lambda_m (Cl^-)$
  • On adding water to a solution of strong electrolyte $\Lambda<em>m$ of strong electrolyte increases because on dilution, ions move away from each other, thus resistance decreases & $\Lambda</em>m$ increases

• Strong electrolyte solution

  • $\Lambda_m ∝$ of strong electrolyte can be determined by the extrapolation method
  • $\Lambda_m^∝$ = molar conductivity at infinite dilution
  • $\Lambda_m^∝$ of an electrolyte is the max possible molar conductivity an electrolyte can show
  • On dilution concentration decreases

• Variation of $\Lambda_m$ of weak electrolyte with dilution

  • Weak electrolytes are not 100% dissociated into ions.
  • On dilution, weak electrolyte dissociates more
  • $\Lambda_m$ for weak electrolyte cannot be determined by the extrapolation method
  • $∝= {K_a \over C}$

KOHLRAUSCH’S LAW OF INDEPENDENT MIGRATION OF IONS

• At infinite dilution, each ion moves with its own characteristic conductance i.e, conductance of an ion is independent of the presence of other ions in its vicinity

APPLICATION OF KOHLRAUSCH’S LAW

• Cal of $\Lambda_m^∝$ of weak electrolytes

  • $\Lambda_m^∝ (HCl)=100$

  • $\Lambda<em>m^∝ (CH</em>3COONa)=200$

  • $\Lambda_m^∝ (NaCl)=80$

  • Cal . $\Lambda_m^∝$ of CH3COOH=?

  • At infinite dilution, a weak electrolyte is 100% dissociated

  • Degree of dissociation (∝) of weak electrolyte at conc ‘C’

    • $∝= {\Lambda<em>{mc} \over \Lambda</em>m^∝}$
    • $\Lambda_{mc} = {K × 1000 \over M}$
    • M= concn = C

• Cal degree of dissociation of 0.1 M crotonic acid (HC) if its sp conductivity is 0.01 Scm-1 .

  • $\Lambda_m^∝(NaC) = 100 Scm^2mol^{-1}$
  • $\Lambda_m^∝(HCl) = 50 Scm^2mol^{-1}$
  • $\Lambda_m^∝(NaCl) = 20 Scm^2mol^{-1}$

• Revise Notes 2 times

• Complete Homework. eg. 2.7, 2.8, 2.9. (H.W)

• NCERT Reading.

• Reading. of Kohlrausch law.

• Best of luck

Topics to be covered

• Batteries
  • Dry cell
  • Pb-storage battery
  • Fuel Cell

Batteries

• Batteries
  • Primary Batteries (Cannot be recharged)
    • Dry Cell
    • Mercury Cell
  • Secondary Batteries (Can be recharged)
    • Lead storage battery
    • Nickel-Cadmium cell
  • Fuel Cells
• The arrangement of two or more galvanic cells connected in the series

Dry Cell (Leclanche Cell)

• Anode : $Zn(s) → Zn^{2+} + 2e^-$
• Cathode : $MnO<em>2 + NH</em>4 ^+ + e^- → MnO(OH) + NH_3$

Dry Cell (Leclanche Cell)

• Manganese is reduced from +4 to +3 state.
• Ammonia produced in the reaction forms a complex with $Zn^{2+}$ to give $[Zn(NH<em>3)</em>4]^{2+}$
• The cell has a potential of nearby 1.5 V.
• Used in transistors and clocks.

Mercury Cell

• Zinc – mercury amalgam as anode and a paste of HgO and carbon as the cathode.
• The electrolyte is a paste of KOH and ZnO.
• Used in hearing aids, watches
• The cell potential is approximately 1.35 V.
  • Anode : $Zn(Hg) + 20H^- → ZnO(s) + H_2O + 2e^-$
  • Cathode : $HgO + H_2O + 2e^- → Hg(I) + 2OH^-$
  • Overall reaction : $Zn(Hg) + HgO(s) → ZnO(s) + Hg(I)$

Lead storage battery

• Negative plates: lead grids filled with spongy lead.
• Positive plates: lead grids filled with PbO,
• 38% sulphuric acid solution

Lead storage battery

• Lead anode and a grid of lead packed with lead dioxide ($PbO_2$) as cathode.

• A 38% solution of $H<em>2SO</em>4$ is electrolyte.

  • Anode : $Pb(s) + SO<em>4 ^{2-} (aq) → PbSO</em>4(s) + 2e^-$
  • Cathode : $PbO<em>2(s) + SO</em>4 ^{2-} (aq) + 4H^+(aq) + 2e^- → PbSO<em>4(s) + 2H</em>2O(I)$
  • Overall reaction: $Pb(s) + PbO<em>2(s) + 2H</em>2SO<em>4(aq) → 2PbSO</em>4(s) + 2H_2O(I)$

• On charging the battery the reaction is reversed and $PbSO<em>4(s)$ on the anode and cathode is converted into Pb and $PbO</em>2$, respectively.

• Anode : $2H<em>2(g) + 4OH^-(aq) → 4H</em>2O(I) + 4e^-$

• Cathode : $O<em>2(g) + 2H</em>2O(I) + 4e^- → 4OH^-(aq)$

• Overall reaction being : $2H<em>2(g) + O</em>2(g) → 2H_2O(I)$

• The cell was used for providing electrical power in the Apollo space programme.

• The water vapors produced during the reaction were condensed and added to the drinking water supply for the astronauts.

• Catalysts like finely divided platinum or palladium metal are incorporated into the electrodes for increasing the rate of electrode reactions.

• Fuel cells produce electricity with an efficiency of about 70% compared to thermal plants whose efficiency is about 40%.

• Question

  • Which of the following cell is a secondary cell ?
    • A Mercury cell
    • B Ni cell
    • C Dry cell
    • D Fuel cell

• Question

  • Which of the following material is not present in a dry cell ?
    • A MnO2
    • B NH4Cl
    • C ZnCl2
    • D KCl

• Question

  • Which of the following material is not present in mercury cell ?
    • A HgO
    • B KOH
    • C Zinc
    • D HgCl2

• Question

  • The approximate voltage of dry cell is:
    • A 2.0
    • B 1.2 V
    • C 6 V
    • D 1.5

• Homework

  • Complete Chap. Revision
  • NCERT Reading
  • Ncert eq., intex Que, Exercise.

FARADAY’S LAWS OF ELECTROPYSIS

• It states that the mass of metal deposited at the cathode is directly proportional to the amount of charge passed through the solution

  • At Cathode $Ag^+(aq) + e^- → Ag (s)$

• ‘Z’ is characteristic of element

  • First LAW $m∝ 𝜃 → 𝑚 = 𝑍𝜃$
  • $𝑚 = 𝑍 × 𝐼 × 𝑇$
  • Z= Electrochemical equivalent

• Physical significance of ‘Z’

  • $𝑚 = 𝑍 × 𝜃$
  • $𝑚 = 𝑍 × 1$
  • If $𝜃 = 1 Coulomb$ then $𝑚 = 𝑍$

• ‘Z’ of an element is equal to the mass of that element deposited by 1 coulomb charge

• Faraday (F) Charge on 1 mole $e^-$

  • $(1F) = 1.6 × 10^{-19} × 6.023 × 10^{23} C$
  • $1𝐹 = 96500 𝐶$

• E= equivalent weight = $\frac{Atomic wt}{valency}$

  • $Ag^{+1} ; Cu^{2+} ; Al^{3+}$

• Cal the charge required to deposit 9 gm $A1^{3+}$.

  • $Al^{+3} +3e^- = Al (s)$
  • 1 mol A1^{3+} -> 3 mole^-
  • 27g -> Al^{3+}
  • Igm Al^{3+} -> 3F charge
  • x gm Al^{3+} -> {3F \over 27} charge

• Q. For how many seconds a current of 10A must be passed through an electrolytic solution order to deposit 10.8 $Ag^+$.

• Cal the Volume of gases evolved at STP if a current of 10A is passed through an electrolytic solution of $Na<em>2SO</em>4(aq)$ for 965 sec.

• Q. An aq. Solution of $AgNO_3$ is electrolyzed using Pt electrodes each having surface area 10 $cm^2$ by a current of 965 Amp for 10 sec. Cal the thickness of Ag deposited on cathode plate (density = 10.8 of Ag gm/cc)

SECOND LAW

• If the same charge is passed through two electrolytic solutions

  • $Eq. wt = {Atomic wt \over valency}$
  • ${M<em>A \over M</em>B} = {E<em>A \over E</em>B}$

• Cal the charge required to deposit 5.6 gm Cd from $CdCl_2$ solution (Atomic wt of Cd = 112)

  • $Cd^{2+} + 2e^- → Cd(s)$
  • Imol Cd - 2F
  • 112gm Cd-) 2x96500
  • 5.6gm cd -) 2x96500 /12 * 56
  • =) 9650 C

• What current (in A) must be passed for 10 min. through a solution of aq $ZnSO<em>4$ solution in order to produce 44.8 L $H</em>2(g)$ at STP

• Revise Notes 2 times

• Complete Homework.

• NCERT Reading

• Best of luck

GALVANIC CELL

• It is an arrangement in which a certain chemical reaction which leads to the generation of electricity

• i.e., it is a device converts chemical energy into electrical energy

• Galvanic cell consists of 2 electrodes:

  • Electrodes: Any arrangement which on combining with a similar arrangement either undergoes oxidation or reduction
    • metal-metal ion electrode
    • gas-gas ion electrode

Galvanic Cell

• Net cell reaction $Zn(s) → Zn^{2+} (g) + 2e^- and Cu^{2+}(g) + 2e^- → Cu(s)$

• $Zn(s) + Cu^{2+}(g) → Zn^{2+} (g) + Cu(s)$

• Salt Bridge

  • Inverted U – shaped tube
  • Electrolyte solution + Agar - Agar
  • Salt bridge maintains electrical neutrality
  • It completes the circuit
    • Zn (s) → $Zn^{2+}$ (g) + 2e-
    • $Cu^{2+}$ (g) + 2e- → Cu(s)

• galvanic cell can also be formed without salt bridge

• external circuit → current direction (cathode to Anode)

• internal circuit → current direction (anode to cathode)

Representation of a galvanic cell

• $Zn (s) | Zn^{2+}(g) | Cu^{2+}(g) | Cu(s)$

• $Zn(s) → Zn^{2+}(g) + 2e^-$

• $Cu^{2+}(g) + 2e^- → Cu(s)$

• $Pt|H_2(g) | H^+ (ax) || Ag^+ (ar) | Ag(s)]$

ELECTROMOTIVE FORCE (EMF)

• EMF is the potential difference in an open circuit
  • $EMF = E<em>{cathode} - E</em>{anode}$
• Electrode potential (reduction potential)

Electrode potential

• It is the potential difference set up between a metal rod & solution of its own ions

• Electrode potential

  • Oxidation potential
  • reduction potential
  • more is the tendency to get oxidized, more will be oxidation potential
  • more is the tendency to get reduced, more will be reduction potential

• $E<em>{oxidation} = -E</em>{reducti}$

• $E_{Zn|Zn^{2+}} = x volt$

• $E_{Zn^{2+}|Zn} = -x volt$

• Two electrode are attached. Cal. EMF of cell

  • $E<em>{A^+|A}= 0.3 V ; E</em>{B^+|B}= 0.7 V$
  • A) 1V
  • B) 0.4 V
  • C) 0.2 V
  • D) none

• Cal. EMF of cell obtained by combining X & Y electrode

  • $E<em>{X|X^+}= 0.8 V ; E</em>{Y|Y^-}= 0.5 W$
  • A) 0.3 V
  • B) 1.3 V
  • C) -0.3 V
  • D) none

• Cal. EMF of cell if both A & B are combined

  • $E<em>{A^-|A}= 0.4 V ; E</em>{B|B^-}= 0.1 V$
  • A) 0.5 V
  • B) 0.3 V
  • C) 0.2 V
  • D) none

• Order of oxidizing nature?

  • $E<em>{A^+|A}= 0.3 V ; E</em>{B^+|B}= 0.7 V ; E_{C^+|C} = -0.3 V$
  • A) A^+ < B^+ < C^+
  • B) C^+ < A^+ < B^+
  • C) A^+ < C^+ < B^+
  • D) none

• Identify the correct statement.

  • $E<em>{X|X^+}= -0.8 V ; E</em>{Y|Y^+}= -0.3 V ; E_{Z|Z^+} = 0.5 V$
  • A) $X^+$ can oxidize both Y & Z
  • B) Z can reduce both $X^+$ & $Y^+$
  • C) X can reduce $Y^+$ but not $Z^+$
  • D) none

• Revise Notes 2 times

• Complete Homework.

• Best of luck

• Factors affecting electrode potential values

  • Concentration of ions
  • Temperature
  • pressure

• Standard conditions for electrodes

  • Concn of ion = 1M
  • T = 298K = 25oC
  • P = 1 atm = 1 Bar

• Standard electrodes

• Calculation of standard electrode potential

  • Standard Hydrogen electrode (SHE/NHE) [Reference for measuring electrode potential values]
    • $E<em>{H</em>2/H^+}^0 = 0$
    • $E<em>{H^+/H</em>2}^0 = 0$

• Q. Cal. $E_{Ag^+/ng}$?

  • $EMF^0 =E<em>{cathode} - E</em>{anode}$

• Electrochemical series

  • It is the ordered arrangement of different elements in increasing or decreasing order of their standard electrode potential
    • E{Na^+/ Na}^0 < E{Mg^{2+}/ Mg}^0 < E{Fe^{2+}/ Fe}^0 < E{Zn^{2+}/ Zn}^0 < E{H^+/ H2}^0 < E{Cu^{2+}/ Cu}^0 < E{Ag^+/ Ag}^0

• $Cu (s) | Cu^{2+}(g) || Ag^+ (g) | Ag (s) ; Cal. E_{cell} ^0$

  • $E_{Cu / Cu^{2+}}^0 = -0.3 V$
  • $E_{Ag^+ / Ag}^0 = 0.8 V$

• Q. Exhibit highest boiling

  • Lighest boiling point?
    • A). 0.1 M Na₂ Soy => i = 2-1 =) 1
    • B). 0.15M C64 1206 => i = 0
    • C). 0.1M Urea =) i = 0
    • D). O. IM KNO3 => i = 1-1 = 0

• $\Delta G = W_{non-PV}$

• $\Delta G = electrical work$

• $\Delta G = -NFe$

• $\Delta G^0 = -NFE^0$

• W=qxv

• -NFE

• Q. $E_{Cu^{2+}/ Cu}^0 = 0.2 V$

  • $E_{Cu^+/ Cu}^0 = 0.7 V$
  • \Cal. E_{Cu^{2+}/ Cu}^0 = ?

• Q. $E<em>{Fe^{2+}/ Fe}^0 = 0.5 V ; E</em>{Fe^{3+}/ Fe}^0 = 0.8 V$

  • \Cal. E_{Fe^{3+}/ Fe^{2+}}^0 = ?

• Cal. Of non-standard electrode potential

  • Nernst Equation
  • $C_{Zn^{2+}}=0.1M$
  • $E_{2n^/2n}=?$
  • $E = E^°- {0.0 \over n} log 10 {1 \over [m+ ]}$

• Cal. Electrode potential of H2 – electrode containing solution having pH = 5 at 298K & 1 atm Pressure

  • $2H^+(aq) + 2e^- → H_2(g)$

• Cal. Non-standard EMF of a galvanic cell $E^0 = 0.8-0.3 0.5 V$

  • $E<em>{Cu^{2+}/ Cu}^0 = 0.3 V ; E</em>{Ag^+/ Ag}^0 = 0.8 V$

• The correct value of cell potential in volt for the reaction that occurs when the following two half cells are connected, is

  • $Fe^{2+} (aq) + 2e^- → Fe(s)$, $E^0 = - 0.44 V$
  • $𝐶𝑟<em>2𝑂</em>7(𝑎𝑞) ^{2−} + 14𝐻^+ + 6𝑒^− → 2𝐶𝑟^{3+} + 7𝐻_2𝑂$, $𝐸^0 = +1.33 𝑉$
  • A) +1.77 V
  • B) +2.65 V
  • C) + 0.01 V
  • D) +0.89 V

• Two half cells reactions are given below.

  • $𝐶𝑜^{3+} + 𝑒 ^− → 𝐶𝑜^{2+}𝐸_{𝐶𝑜^{2+}/ 𝐶𝑜^{3+}}^0 = −1.81 𝑉$
  • $2𝐴𝑙^{3+} + 6𝑒^− → 2𝐴𝑙 𝑠 E_{𝐴𝑙 /𝐴𝑙^{3+}}^0 = +1.66 𝑉$
  • The standard EMF of a cell with a feasible redox reaction will be:
    • A) -3.47 V
    • B) +7.09 V
    • C) +0.15 V
    • D) +3.47 V

• A button cell used in watches functions as following $Zn(s) + Ag<em>2O(s) + H</em>2O (l) → 2Ag (s) + Zn^{2+}(aq) + 2OH^-(aq)$. If half cell potentials are:

  • $Zn^{2+}(aq) + 2e^- ; Eo = -0.76 V$
  • $Ag<em>2O(s) + H</em>2O (l) + 2e^- → 2Ag(s) + 2OH^-(aq). E^0 = 0.34 V$. The cell potential will be:
    • A) 1.10 V
    • B) 0.42 V
    • C) 0.84 V
    • D) 1.32 V

• Standard electrode potential for the cell with cell reaction $Zn(s) + Cu^{2+} (aq) → Zn^{2+}(aq) + Cu(s)$ is 1.1 V. Calculate the standard Gibbs energy change for the cell reaction. (Given $F = 96487 C mol^{-1}$)

  • A) -200.27 J mol-1
  • B) -200.27 kJ mol-1
  • C) -212.27 kJ mol-1
  • D) -212.27 J mol-1

• Find the emf of the cell in which the following reaction takes place at 298 K

  • $Ni(s) + 2Ag^+(0.001M) → Ni^{2+} (0.001M) + 2Ag(s)$
  • (Given that $E^0 cell=10.5 V, {2.303RT \over F} = 0.059$ at 298 K)
    • A) 1.05 V
    • B) 1.0385 V
    • C) 1.385 V
    • D) None of these

• Revise Notes 2 times

• Complete Homework.

• Best of luck

• A button cell used in watches functions as following $Zn(s) + Ag<em>2O(s) + H</em>2O (l) → 2Ag (s) + Zn^{2+}(aq) + 2OH^-(aq)$. If half cell potentials are:

  • $Zn^{2+}(aq) + 2e^- ; Eo = -0.76 V$
  • $Ag<em>2O(s) + H</em>2O (l) + 2e^- → 2Ag(s) + 2OH^-(aq). E^0 = 0.34 V$. The cell potential will be:
    • A) 1.10 V
    • B) 0.42 V
    • C) 0.84 V
    • D) 1.32 V

• Standard electrode potential for the cell with cell reaction $Zn(s) + Cu^{2+} (aq) → Zn^{2+}(aq) + Cu(s)$ is 1.1 V. Calculate the standard Gibbs energy change for the cell reaction. (Given $F = 96487 C mol^{-1}$)

  • A) -200.27 J mol-1
  • B) -200.27 kJ mol