1. Calculate the e.m.f of zinc-silver cell at 25°C when [Zn2+] = 0.10 M and [Ag+] = 10 M. (E0cell at25°C= 1.56 volts). Or Calculate the e.m.f of the zinc-silver cell at 25°C. When [Zn2+] = 0.10 M and [Ag+] = 10 M, cell reaction is 2Ag+ + Zn ⇌ 2Ag + Zn2+; E°cell at 25° C = 1.56 V. Or Calculate the e.m.f of the cell having the cell reaction 2Ag+ +Zn ⇌ 2Ag + Zn2+; E°cell = 1.56 V at 25°C when concentration of Zn2+ = 0.1 M and Ag+ = 10 M in the solution.
The cell reaction in the zinc - silver cell would be: 2Ag+ + Zn ⇌ 2Ag + Zn2+
The Nernst equation for the above cell reaction may be written as:
Ecell = Eocell – RT / nF ln [Ag]2 [Zn2+] / [Ag+]2 [Zn]
(Since concentrations of solids are taken as unity i.e., [M] = 1)
Ecell = Eocell – RT / nF ln [Zn2+] / [Ag+]2
Substituting the various values in Nernst equation, we have
Ecell = 1.56 – (2.303 x 8.314 x 298 / 2 x 96495) log 0.1 / 102
= 1.56 – 0.02957 log 10–3
= 1.56 + 3(0.02957) = 1.56 + 0.0887
= 1.6487 volts
2. Calculate the e.m.f of the cell Zn | Zn2+(0.001 M) ||Ag+(0.1M) | Ag E°Ag/Ag+ = + 0.80 V. E°Zn/Zn2+= - 0.76V.
Step 1: Write the half-cell reactions of the anode and the cathode. Then add the anode and cathode half reactions to obtain the cell reaction and the value of Eo cell.
At Cathode: 2Ag+ + 2e– → 2Ag Eo = + 0.80 V
At Anode: Zn → Zn2+ + 2e– Eo = – 0.76 V
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Cell reaction: 2 Ag+ + Zn ⇌ Zn2+ + 2Ag Eo = 1.56 V
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Step 2: K for the cell reaction = [Zn2+] / [Ag+]2
Substituting the values
n = 2 electrons
Eocell = 1.56 V
[Zn2+] = 0.001 M = 10–3 M
[Ag+] = 0.1 M = 10–1 M
in the Nernst equation and solving for Ecell, we have
Ecell = Eocell – 0.0591 / n log K
= 1.56 – (0.0591 / 2) log [Zn2+] / [Ag+]2
= 1.56 – 0.0591 / 2 log [10–3] / [10–1]2
= 1.56 – 0.02955 x log 10–1
= 1.56 + 0.02955
= 1.58955 V
3. Calculate the equilibrium constant for the following cell reaction 2Ag+ + Zn ⇌ Zn2+ + 2Ag E°Ag+/Ag = + 0.80 V. E°Zn2+/ Zn = - 0.76V.
Step 1: Write the half-cell reactions of the anode and the cathode. Then add the anode and cathode half reactions to obtain the cell reaction and the value of Eocell.
At Cathode: 2Ag+ + 2e– → 2Ag Eo = + 0.80 V
At Anode: Zn → Zn2+ + 2e– Eo = – 0.76 V
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Overall Cell reaction: 2 Ag+ + Zn ⇌ Zn2+ + 2Ag Eo = 1.56 V
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Step 2: Substitute the values Ecell = 0, n = 2 & Eocell = 1.56 V in the Nernst equation at equilibrium
At equilibrium Cell potential, Ecell = 0 and Reaction quotient, J = Equilibrium constant, K
At equilibrium Cell potential, Ecell = 0 and Reaction quotient, J = Equilibrium constant, K
∴ 0 = Eocell – 0.0591 / nF log K
The Nernst equation can be rearranged as:
nFEocell = 0.0591 log K
∴ log K = nEocell / 0.0591
log K = 2 × 1.56 / 0.0591
= 52.7919
K = Antilog 52.7919
K = 6.19 x 1052
4. Calculate the standard e.m.f and standard free energy change of the following cell: Zn | Zn2+ || Cu2+ | Cu E°Zn2+/ Zn = – 0.762 V, E°Cu2+/ Cu = + 0.337 V.
Eocell = EoR – EoL
Eocell = EoR – EoL
= + 0.337 – (– 0.762)
= + 1.099 V
Eocell is + ve. ∴ ΔGo = – ve.
∴ ΔGo = – n FEocell
n = 2 electrons
∴ ΔGo = – 2 x 96495 x 1.099
= – 212096 Joules
= – 212.096 kJ.
5. Derive Nernst equation (of reversible cell.)
The reaction occurring in a reversible cell is represented by the equation
A + B ⇌ C + D
The decrease in free energy, – ΔG, accompanying the process is given by
– ΔG = – ΔGo – RT ln J
Where,
– ΔGo is the decrease in free energy accompanying the same process when all the reactants and products are in their standard states of unit activity.
J stands for the reaction quotient of the activities of the products and reactants at any given stage of the reaction.
Substituting the value of J, we have
– ΔG = – ΔGo – RT ln aC x aD / aA x aB
If E is the E.M.F. of the cell in volts and the cell reaction involves the passage of ‘n’ faradays i.e., nF coulombs, the electrical work done by the cell is in nFE volt-coulombs or Joules.
Hence free energy decrease of the system, – ΔG, is given by the expression
– ΔG = nFE
nFE = – ΔGo – RT ln aC x aD / aA x aB
= nFEo – RT ln aC x aD / aA x aB
E = Eo – RT / nF ln aC x aD / aA x aB is known as the Nernst equation
Where,
Eo is the E.M.F. of the cell in which the activity, or as an approximation, the concentration of each reactant and each product of the cell reaction is equal to unity. Eo is known as the standard E.M.F. of the cell.
Replacing activities by concentrations the Nernst equation may be written as
E = Eo – RT / nF ln [C][D] / [A][B]
Replacing [C][D] / [A][B] as equal to K, the equilibrium constant in the molar concentration units,
E = Eo – RT / nF ln K.
E = Eo – 2.303 RT / nF log K
Where,
Eo = Standard electrode potential
R = Gas constant,
T = Kelvin temperature
n = Number of electrons transferred in the half-reaction
F = Faraday of electricity
K = Equilibrium constant for the half-cell reaction as in equilibrium law.
Substituting the values of R (8.314 J K-1 mol-1), F (96,495 coulombs) and T at 25oC (273 + 25 = 298 K), the quantity 2.303 RT / F comes to be 0.0591.
Thus the Nernst equation can be written in its simplified form as
E = Eo – 0.0591 / n log K
6. Determine the standard e.m.f of the cell and standard free energy change of the cell reaction. Zn, Zn2+ || Ni2+, Ni. The standard reduction potentials (E°) of Zn2+, Zn and Ni2+, Ni half cells are – 0.76 V and – 0.25 V respectively.
Eocell = EoR – EoL
Eocell = EoR – EoL
= – 0.25 – (– 0.76)
= + 0.51 V
Eocell is + ve. ∴ ΔGo = – ve.
∴ ΔGo = – n FEocell
n = 2 electrons
∴ ΔGo = – 2 x 96495 x 0.51
= – 98425 Joules
= – 98.43 kJ.
7. How is a Standard Hydrogen Electrode (SHE) constructed? Explain its function.
The standard hydrogen half-cell or Standard Hydrogen Electrode (SHE) is selected for coupling with the unknown half-cell. It consists of a platinum electrode immersed in a 1 M solution of H+ ions maintained at 25oC. Hydrogen gas at one atmosphere enters the glass hood and bubbles over the platinum electrode. The hydrogen gas at the platinum electrode passes into solution, forming H+ ions and electrons.
The emf of the standard hydrogen electrode is arbitrarily assigned the value of zero volts. So, SHE can be used as a standard for other electrodes.
The half-cell, whose potential is desired, is combined with the hydrogen electrode and the emf of the complete cell determined with a voltmeter. The emf of the cell is the emf of the half-cell.
For example, it is desired to determine the emf of the zinc electrode, Zn | Zn2+. It is connected with the SHE. The complete electrochemical cell may be represented as:
Zn | Zn2+ || H+ | H2 (1 atm), Pt
The emf of the cell has been found to be – 0.76 V which is the emf the zinc half-cell. Similarly, the emf of the copper electrode, Cu2+ | Cu can be determined by pairing it with the SHE when the electrochemical cell can be represented as:
Pt, H2 (1 atm) | H+ || Cu2+ | Cu
The emf of this cell has been determined to be 0.34 V which is the emf of the copper half-cell.
Eocell = EoCu/Cu2+ – EoSHE
= 0.34 – Zero
= 0.34 V
The two situations are explained as follows:
When it is placed on the right-hand side of the zinc electrode, the hydrogen electrode reaction is
2H+ + 2e– → H2
The electrons flow to the SHE and it acts as the cathode.
When the SHE is placed on the left hand side, the electrode reaction is
H2 → 2H+ + 2e–
The electrons flow to the copper electrode and the hydrogen electrode as the anode. Evidently, the SHE can act both as anode and cathode and, therefore can be used to determine the emf of any other half-cell electrode (or single electrode).
8. The e.m.f of the half cell Cu2+(aq) / Cu(s) containing 0.01M Cu2+ solution is + 0.301 V. Calculate the standard e.m.f of the half cell.
e.m.f of the half cell, ECu2+/ Cu = 0.301 V
Cu2+ (aq) + 2 e─ ¾® Cu (s)
ஃ n = 2 electrons
ஃ n = 2 electrons
[Cu2+] = 0.01M = 10-2 M
Standard e.m.f of the half cell, EoCu2+/ Cu = ?
ECu2+/ Cu = EoCu2+/ Cu – 0.0591 / 2 log 1 / [Cu2+]
0.301 = EoCu2+/ Cu + 0.0591 / 2 log [Cu2+]
Or EoCu2+ / Cu + 0.0591 / 2 log [Cu] / [Cu2+]
Or EoCu2+ / Cu + 0.0591 / 2 log [Cu] / [Cu2+]
0.301 = EoCu2+/ Cu + 0.0591 / 2 log 10–2
Or EoCu2+ / Cu + 0.0591 / 2 log 1 / 10 2 M
Or EoCu2+ / Cu + 0.0591 / 2 log 1 / 10 2 M
0.301 = EoCu2+/ Cu + 0.0591/2 x (– 2)
0.301 = EoCu2+/ Cu – 0.0591
EoCu2+/ Cu = 0.301 + 0.0591
= 0.3601 V
9. Write an account on Cell terminology.
1. Current is the flow of electrons through a wire or any conductor.
2. Electrode is a metallic rod / bar / strip which conducts electrons into and out of a solution.
3. Anode is the electrode at which oxidAtion occurs. It sends electrons into the outer circuit. It has negative charge and is shown as (–) in cell diagrams.
4. Cathode is the electrode at which reduCtion occurs. It receives electrons from the outer circuit. It has a positive charge and is shown as (+) in the cell diagrams.
5. Electrolyte is the salt solution in a cell.
6. Anode compartment is the compartment of the cell in which oxidation half-reaction occurs. It contains the anode.
7. Cathode compartment is the compartment of the cell in which reduction half-reaction occurs. It contains the cathode.
8. Half-cell - Each half of an electrochemical cell, where oxidation occurs and the half where reduction occurs, is called the half cell.
10. Derive the relation between e.m.f and free energy. Or Establish a relation between free energy and e.m.f Or Write a brief account on the relation between EMF and free energy.
When a cell produces a current, the current can be used to do work – to run a motor, for instance. Thermodynamic principles can be employed to derive a relation between electrical energy and the maximum amount of work, Wmax, obtainable from the cell. The maximum amount of work obtainable from the cell is the product of charge flowing per mole and maximum potential difference, E, through which the charge is transferred.
Wmax = – n FE ... (1)
Where,
n is the number of moles of electrons transferred and is equal to the valence of the ion participating in the cell reaction.
F stands for Faraday and is equal to 96,495 coulombs
E is the emf of the cell.
According to thermodynamics, the maximum work that can be derived from a chemical reaction is equal to the free energy (ΔG) for the reaction,
Wmax = ΔG ... (2)
Therefore, from (1) and (2), we can write
ΔG = – n FE ... (3)
Thus only when E has a positive value, ΔG value will be negative and the cell reaction will be spontaneous and the e.m.f. of the cell can be measured.
Here E refers to the Ecell.
Thus, the electrical energy supplied by the cell is (nFE) equal to the free energy decrease (– ΔG) of the cell reaction occurring in the cell.
11. Describe Daniel cell.
Daniel cell or a galvanic cell is an example of electrochemical cell.
The overall reaction taking place in the cell is the redox reaction given as
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
This overall reaction is made of the summation of two half reactions such as oxidation half reaction and reduction half reaction.
The oxidation half reaction occurring at the zinc electrode in contact with the aqueous electrolyte containing Zn2+, accumulates the electrons at the zinc rod.
Zn(s) → Zn2+(aq) + 2e–
The reduction half reaction occurring at the copper electrode in contact with the aqueous electrolyte containing Cu2+ ions receives the electrons from the zinc electrode when connected externally, to produce metallic copper according to the reaction as,
Cu2+ + 2e– → Cu(s)
The decrease in the energy which appears as the heat energy when a zinc rod is directly dipped into the zinc sulphate solution, is converted into electrical energy when the same reaction takes place indirectly in an electrochemical cell.
The Daniel cell is also called as the voltaic cell. However for continuous supply of current for a long period, the two half cells each comprising the metal electrode and its aqueous electrolyte kept in separate containers and can be connected externally as below:
When the cell is set up, electrons flow from zinc electrode through the wire to the copper cathode. As a result, zinc dissolves in the anode solution to form Zn2+ ions. The Cu2+ ions in the cathode half cell pick up electrons and are converted to Cu atoms on the cathode.
12. How is e.m.f of a half cell determined?
The emf of the unknown half-cell Eo is calculated by constructing a cell with standard half-cell.
Emeasured = ER – EL
If the standard half-cell acts as anode, the equation becomes ER = Emeasured (∵ EL = 0)
If standard half-cell acts as cathode EL = – Emeasured (∵ ER = 0)
The standard hydrogen half-cell or Standard Hydrogen Electrode (SHE) is selected for coupling with the unknown half-cell.
The emf of the standard hydrogen electrode is arbitrarily assigned the value of zero volts. So, SHE can be used as a standard for other electrodes.
The SHE can act both as anode and cathode and, therefore can be used to determine the emf of any other half-cell electrode (or single electrode).
13. Write IUPAC representation of a cell. Or Write notes on IUPAC convention of representation of a cell. Or Write the IUPAC conventions for writing cell diagram with examples.
A cell diagram is an abbreviated symbolic depiction of an electrochemical cell.
An electrochemical cell consists of two half-cells. Each half-cell is made of a metal electrode in contact with metal ion in solution.
IUPAC recommended the following conventions for writing cell diagram. We will illustrate these with reference to Zinc-Copper cell.
1. A single vertical line ( | ) represents a phase boundary between metal electrode and ion solution (electrolyte).
Zn | Zn2+ Cu2+ | Cu
Phase Boundary
Anode half-cell Cathode half-cell
It may be noted that the metal electrode in anode half-cell is on the left, while in cathode half-cell it is on the right of the metal ion.
2. A double vertical line ( || ) represents the salt bridge, porous partition or any other means of permitting ion flow while preventing the electrolyte from mixing.
3. Anode half-cell is written on the left and cathode half-cell on the right.
4. In the complete cell diagram, the two half-cells are separated by a double vertical line (salt bridge) in between. The zinc-copper cell can now be written as
Zn | Zn2+ || Cu2+ | Cu
Anode half-cell Cathode half-cell
5. The symbol for an inert electrode, like the platinum electrode is often enclosed in a bracket. Example
Mg | Mg2+ || H+ | H2 (Pt)
Anode half-cell Cathode half-cell
(6) The value of emf of a cell ( E ) is written on the right of the cell diagram. Thus a zinc-copper cell has emf 1.1 V and is represented as
Zn | ZnSO4 || CuSO4 | Cu E = + 1.1 V
Direction of electron flow →
If the emf acts in the opposite direction through the cell circuit it is denoted as a negative value.
Cu | CuSO4 || ZnSO4 | Zn E = – 1.1 V
← Direction of electron flow
The negative sign also indicates that the cell is not feasible in the given direction and the reaction will take place in the reverse direction only. The overall cell reaction for E = – 1.1 V of the Daniel cell is
Cu(s) + Zn(aq)2+ ⇌ Cu(aq)2+ + Zn(s)
The reversal of the cell current is accompanied by the reversal of direction of the cell reaction.
14. Write notes on single electrode potential.
An electrochemical cell consists of two half-cells. With an open-circuit, the metal electrode in each half-cell transfers its ions into solution. Thus an individual electrode develops a potential with respect to the solution. The potential of a single electrode in a half-cell is called the Single electrode potential. Thus in Daniel cell in which the electrodes are not connected externally, the anode Zn / Zn2+ develops a negative charge and the cathode Cu / Cu2+, a positive charge. The amount of the charge produced on an individual electrode determines its single electrode potential.
The single electrode potential of a half-cell depends on:
a. Concentration of ions in solution
b. Tendency to form ions and
c.Temperature.
Standard emf of a cell
The emf generated by an electrochemical cell is given by the symbol E. It can be measured with the help of a potentiometer. When the emf of a cell is determined under standard conditions, it is called the standard emf. The standard emf may be defined as the emf of a cell with 1 M solutions of reactants and products in solution measured at 25oC. Standard emf of a cell is represented by the symbol Eo. For gases 1 atm. pressure is a standard condition instead of concentration. For Zn-Cu voltaic cell, the standard emf, Eo is 1.10V.
Zn | Zn2+(aq, 1M) || Cu2+(aq, 1M) | Cu Eo = 1.10 V
Applications of Single electrode potential
1. Predicting Cell EMF:
The standard emf Eo, of a cell is the standard reduction potential of right-hand electrode (cathode) minus the standard reduction potential of the left-hand electrode (anode).
That is, Eocell = Eoright – Eoleft
That is, Eocell = Eoright – Eoleft
= Cathode potential – Anode potential
2. Predicting Feasibility of Reaction:
The feasibility of a redox reaction can be predicted with the help of the electrochemical series. The net emf of the cell reaction, Ecell, can be calculated from the expression
Eocell = Eocathode – Eoanode
In general, if Eocell = + ve, the reaction is feasible
Eocell = – ve, the reaction is not feasible.
3. Metal displacement: (Predicting whether a metal will displace another metal from its salt solution or not)
The metals near the bottom of the electrochemical series are strong reducing agents and are themselves oxidised to metal ions. On the contrary, the metal lying higher up in the series are strong oxidizing agents and their ions are readily reduced to the metal itself.
Eg., Zinc lying down below the series is oxidised to Zn2+ ion, while copper which is higher up in the series is produced by reduction of Cu2+ ion.
In general we can say that a metal lower down the electrochemical series can precipitate the one higher up in the series.
Eg., Silver cannot precipitate Cu from CuSO4 solution, since both metals have positions higher up in the series and are strong oxidising agents.
4. Hydrogen displacement: (Predicting whether a metal will displace hydrogen from a dilute acid solution)
Any metal above hydrogen in the electrochemical series is a weaker reducing agent than hydrogen and will not convert H+ to H2. This explains why Zn lying below hydrogen reacts with dil. H2SO4 to liberate H2, while Cu lying above hydrogen does not react.
[The electrochemical series / reactivity series is so-called because it is a method of classifying elements according to their chemical reactivity.]
15. With the help of the
electrochemical series, how will you predict whether a metal will displace another
metal from its salt solution or not.
The metals near the bottom of the
electrochemical series are strong reducing agents and are themselves oxidised
to metal ions. On the contrary, the metal lying higher up in the series are
strong oxidizing agents and their ions are readily reduced to the metal itself.
Eg., Zinc lying down below the series is
oxidised to Zn2+ ion, while copper which is higher up in the
series is produced by reduction of Cu2+ ion.
Zn →
Zn2+ + 2e–
Cu2+ +
2e– →
Cu ↓
Thus when zinc is
placed in CuSO4 solution, Cu metal gets precipitated
In general we can say that a metal lower
down the electrochemical series can precipitate the one higher up in the
series.
Eg., Silver
cannot precipitate Cu from CuSO4 solution, since both metals have positions
higher up in the series and are strong oxidising agents
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