Electrochemical Cell And Nernst Equation

Electrochemical cell:

It is a device or arrangement by which we can convert chemical energy into electrical energy i.e. chemical energy produced as a result of redox chemical reaction of cell.

Some major points regarding electrochemical cell:

(i) Electrochemical cell consists of two half – cells. The half – cell in which oxidation occurs is called oxidation half – cell and the another half cell in which reduction occurs is called reduction half cell.

(ii) Electrons flow from anode to cathode in the external circuit.

(iii) Chemical energy is converted into electrical energy.

(iv) The net reaction is the sum of two half – cell reactions.

Oxidation half reaction $M_{1(s)}\to M{{}_1^{++}}_{(aq)}+2e^{-}$

$\text{Reduction half reaction }M{{}_2^{++}}_{(aq)}+2e^{-}\to M_{2\left(s\right)}$

$M_{1\left(s\right)}+M{{}_2^{++}}_{\left(aq\right)}\to M{{}_1^{++}}_{\left(aq\right)}+M_{2\left(s\right)}$

(iv) Two half – cells are connected by the salt bridge and completes the cell circuit.

(v) Salt bridge prevents transference or diffusion of the solutions from one half – cell to the other. It also helps to maintain the electrical neutrality of half cells.

(vi) Salt bridge is represented by a broken line or two parallel vertical lines in a cell reaction.

$M/M{{}_1^{++}}_{(aq)}||M{{}_2^{++}}_{(aq)}/M_2$

(vii) To sum up:

Left $Halfelectrode\stackrel{Electronsflow}{\to }Righthalfelectrode$

Metal / Metal ion (conc.)||Metal ion (conc.) / Metal

Oxidation occursSalt bridgeReduction occurs

Anode Cathode

Negative pole Positive pole

ELECTRODE POTENTIAL

(a) Oxidation potential:

Potential developed on anode electrode due to oxidation process is called oxidation potential.

$M\to M^{n+}+ne$

(b) Reduction potential:

Potential developed on cathode electrode due to reduction process is called reduction potential. $M^{n+}+ne\to M$

Emf of the cell = oxidation potential of anode + Reduction potential of cathode

Reference electrode (Standard Hydrogen Electrode):

By this device, we measure, the standard electrode potential of half cell. It consists of a small platinum strip coated with platinum black so as to absorb H₂ gas. A platinum wire is welded to the platinum strip and sealed in a glass tube. The entire arrangement dipped into 1 M HCl solution at 298 K and specified pressure of H₂ gas.

In this arrangement oxidation and reduction process occurs at same electrode and resultant potential becomes equal to zero.

$H_2\to 2H^{+}+2e^{-}(oxidationprocess)$

$2H^{+}+2e^{-}\to H_2(\mathrm{Re}ductionprocess)$

(i) Measurement of electrode potential of $M_1/M_1^{n+}$ electrode, which placed above than hydrogen in electro – chemical series.

(ii) Measurement of electrode potential of $M_2/M_2^{n+}$ electrode, which placed below than hydrogen in electro – chemical series.

Some other reference electrode:

(i) Calomel electrode – Hg, Hg₂Cl₂ / $C{l^{-}}_{(aq)}$

(ii) Silver – silver chloride electrode – Ag, AgCl / $C{l^{-}}_{(aq)}$

Various types of half cells:

Illustration 1. Hydrogen gas will reduces

(A) heated cupric oxide (B) heated ferric oxide

(C) heated stannic oxide (D) heated aluminium oxide

Solution: The standard reduction potential of Copper is greater than the standard reduction potential of hydrogen. So hydrogen gas reduces cupric oxide as

CuO + H₂ $\to$ Cu + H₂O

Similarly, hydrogen reduces ferric oxide to ferous oxide

Fe₂O₃ + 3H₂ $\to$ 2FeO + 3H₂O

Hence, (A) & (B) is correct

Emf of cell:

The Nernst Equation:

For a reversible reaction,

$\alpha A+\beta B\rightleftharpoons \gamma C+\delta D$

Reaction Quotient / Equilibrium constant $K=\frac{{\left[D\right]}^{\lambda }}{{\left[A\right]}^{\alpha }}\frac{{\left[D\right]}^{\delta }}{\left[B\right]{}^{\beta }}.$

During cell reaction, $\Delta G=$change in Gibbs free energy which is equal to the available work done by cell.

According to thermodynamic consideration,

$\Delta G=\Delta G^o+RT\ell nQ$

where, $\Delta G=$Change in Gibbs free energy

$\Delta G^o=$Standard change in Gibbs free energy

Q = Reaction quotient

or, $-nFE=-nF{E}^o+2.303RT{\log }_{10}Q$

or, $E=E^o-\frac{2.303RT}{nF}\log Q$

or,$E=E^o-\frac{0.0591}{n}\log Q$If R = 8.314 J $k^{-1}mo{l}^{-1}$T = 298 KF = 96500or,$E_{cell}=E_{cell}^o-\frac{0.0591}{n}\log \frac{[\mathrm{Pr}oducts]}{[\mathrm{Re}ac\tan ts]}$$E_{cell}^o=E{{}_{ox}^o}_{(anode)}+E{{}_{\mathrm{Re}d}^o}_{(cathode)}$;;;;;;$=E_{\mathrm{Re}d(cathode)}^o-E_{\mathrm{Re}d(anode)}^o$;;;where $E_{(ox)}^o=-E_{\mathrm{Re}d}^o$ (For same electrode);;;;;;$=E_{RHS}^o-E_{LHS}^o$

Illustration 2.;;;For the cell Tl || Tl⁺ (0.001M) || Cu²⁺ (0.1M) | Cu. E_Cell at 25°C is. 83V which can be increased;;;;;;;;;;;;;;;;;;;;;;;(A) by increasing [Cu²⁺];;;;;;;;;;;;;;;;;;;;;;;;;(B) by increasing [TI⁺];;;;;;;;;;;;;;;;;;;;;;;(C) by decreasing [Cu²⁺];;;;;;;;;;;;;;;;;;;;;;;;(D) by decreasing [TI⁺]Solution:;;;;;;;2Tl + Cu⁺² $\to$ 2Tl⁺ + Cu;;;;;;;;;;;;;;;;;;;;;;$E_{cell}=E_{cell}^0-\frac{0.0591}{2}\log \frac{{\left[T{I}^{+}\right]}^2}{\left[C{u}^{+2}\right]}$;;;;;;;;;;;;;;;;;;;;;;Hence, (A) & (D) is correct.

Applications of Nernst equation:(a);Determination of equilibrium constant for cell reaction:;;;;;If $E_{cell}=E_{cell}^o-\frac{0.0591}{n}\log Q$ (Q = Reaction Quotient);;;;;At equilibrium, $E_{cell}=0andQ=K_{eq}$;;;;;$\therefore E_{cell}^o=\frac{0.0591}{n}\log K_{eq}$;;;;;$\therefore \log K_{eq}=\frac{E_{cell}^o\times n}{0.0591}$;;;;$\therefore K_{eq}=Anti\log \left[\frac{E_{cell}^0\times n}{0.0591}\right]$;;;;;(b);Determination of solubility product:;;;;;If cell reaction is;;;;;$Ag/A{g}^{+},saturatedKCl//A{g}^{+}(2M)/Ag$;;;;;$\therefore E_{cell}=E_{cell}^o-\frac{0.0591}{1}\log \frac{[A{g}^{+}]}{[x]}$;;;;;$\therefore E_{cell}=\frac{-0.0591}{1}\log \frac{[A{g}^{+}]}{[x]}where{E}_{cell}^o=0$for concentration cell;;;;;$\therefore E_{cell}=0.0591\log \frac{x}{[A{g}^{+}]}$;;;;;Value of $[A{g}^{+}]$ can be calculated and;;;;;;;;;;;$K_{sp}(AgCl)=[A{g}^{+}][C{l}^{-}]={[A{g}^{+}]}^2$

Illustration 3.;Find the solubility product of a saturated solution of Ag₂CrO₄ in water at 298K if the emf of the cell:;;;;;;;;;;;;;;;;;;;;;;;Ag | Ag⁺;(sat Ag₂CrO₄ solution) || Ag⁺;(0.1 M) | Ag is 0.164 at 298KSolution:;;;;;;;For cell Ag | Ag⁺;(saturated solution of Ag₂CrO₄) || Ag⁺;(0.1 M) | Ag;;;;;;;;;;;;;;;;;;;;;;;E_cell = 0.164V;;;;;;;;;;;;;;;;;;;;;;;E_cell = E⁰_cell –$\frac{0.059}{1}\log \frac{{[A{g}^{+}]}_{LHS}}{{[A{g}^{+}]}_{RHS}}$;;;;;;;;;;;;;;;;;;;;;;;$\Rightarrow$ 0.164 = $E_{Ag/A{g}^{+}}^0+E_{A{g}^{+}/4}^0-\frac{0.059}{1}\log \frac{{[A{g}^{+}]}_{LHS}}{0.1}$;;;;;;;;;;;;;;;;;;;;;;;$\therefore$ [Ag⁺]_LHS = 1.66 × 10^–4M;;;;;;;;;;;;;;;;;;;;;;;K_sp for Ag₂CrO₄;$\rightleftharpoons$;;2Ag⁺ + CrO₄^2–;;;;;;;;;;;;;;;;;;;;;;;K_sp;= [Ag⁺]² [CrO₄^2–] = [1.66 × 10^–4]² $\left[\frac{1.66\times {10}^{-4}}{2}\right]$;;;;;;;;;;;;;;;;;;;;;;;$\therefore$ K_sp = 2.287 × 10^–12 mol³ liltre^–3

Ionisation constant and degree of dissociation of weak acid or weak base:If cell reaction is Pt(H₂) / H⁺, HA(C₁) // Ag⁺ (C₂) / Ag$E_{cell}=E_{cell}^o-\frac{0.0591}{1}\log \frac{[H^{+}]}{[A{g}^{+}]}$$E_{cell}=E_{cell}^o-\frac{0.0591}{1}\log \frac{{(K_{a}C)}^{1/2}}{C_2}$By this expression, we can calculate the value of K_a for acid.If value of K_a and C are known, then we can calculate the value of $\alpha =\sqrt{\frac{K_a}{C}}$Determination of pH of a solution:If cell reaction is:$Pt(H_2),(1atm)/H^{+}(pH=x)//C{l}^{-}(1M)/H{g}_{2}C{l}_2,Hg$$E_{cell}=E_{cell}^o-\frac{0.0591}{2}\log \frac{{[H^{+}]}^2\times {[C{l}^{-}]}^2}{H_{2(g)}\times H{g}_{2}C{l_2}_{(s)}}$;;;;;$\therefore E_{cell}=E_{cell}^o-\frac{0.0591}{2}\log {\left[H^{+}\right]}^2$or,;$E_{cell}=E_{cell}^o+0.0591pH$or, $pH=\frac{E_{cell}-E_{cell}^o}{0.0591}$

Illustration4.;;;;By how much is the oxidizing power of the MnO₄^–/Mn²⁺ couple be decreased if the H⁺ concentration is decreased from 1 M to 10^–4 M at 25°C? Assume other species have no change in concentration.Solution:;;;;;MnO₄^– (in acidic medium) is oxidizing agent.;;;;;;;;;;;;;;;;;;;;;MnO₄^– + 8H⁺ + 5e^–$\to$ Mn²⁺ + 4H₂OE – E⁰= $-\frac{0.059}{2}\log \frac{[M{n}^{2+}][H_{2}O]}{[Mn{O}_4^{-}]{[H^{+}]}^8}$ =;$-\frac{0.059}{5}\log \frac{1}{{\left({10}^{-4}\right)}^8}$ = – 0.38 VThus MnO₄⁺/Mn²⁺couple will move to position of less oxidizing power by 0.38 V.

Illustration 5.Graph between E_cell R log $\frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]}$ was linear with intercept on E_cell^;axis 1.1 V. Calculate E_cell forZn | Zn²⁺(0.1 M) || Cu²⁺(0.01 M) +| Cu.Solution: Cell reaction is Zn + Cu²⁺$\to$ Zn²⁺;+ Cu;;;;;;;;;;;;;;;;;;;;;E_cell= E⁰_cell xx;;;;$0.0\frac{0.059}{n}.\log \frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]}$;;;;;.;;;;;;;;;;;…(1);;;;;;;;;;;;;;;;;;;;;We can compare equation (1) with y = mx + c;;;;;;;;;;;;;;;;;;;;;$\therefore$;;;C = E⁰_cell = 1.1 V;;;;;;;;;;;;;;;;;;;;;;$\therefore$;;;E_cell = E⁰_cell –$\frac{0.0591}{2}\log \frac{[Z{n}^{2+}]}{[C{u}^{2+}]}$= 1.1 –$\frac{0.059}{2}$log10;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;= 1.1 – 0.295 = 1.0705 V

DETERMINATION OF THERMODYNAMICAL DATA$\Delta G=\Delta H+T{\left[\frac{d(\Delta G)}{dt}\right]}_p$ (Gibbs – Helmholtz equation)$\Rightarrow -nFE=\Delta H-nFT\left[\frac{d_{E_{cell}}}{dT}\right]$or, $\Delta H=-nF{E}_{cell}+nFT\left[\frac{{d_E}_{{}_{cell}}}{dT}\right]$or, $\Delta H=-nF\left\{E_{cell}-T\left[\frac{{d_E}_{{}_{cell}}}{dT}\right]\right\}$$\left[\frac{d_{E_{cell}}}{dT}\right]=\frac{\Delta H}{nFT}+\frac{E_{cell}}{T}$(Temperature coefficient of the emf of the cell)Illustration 6.Calculate the standard free energy change in kJ for the reaction:$C{u}^{+}+I^{-}\to CuI$Given:$CuI+e\to Cu+I^{-}{E}^0=-0.17V$$C{u}^{+}+e\to Cu{E}^0=0.53V$(A)$-67.55$(B)135.1(C)1.78V(D)-1.75VSolution:$CuI+e\to Cu+I^{-}$$E_1^0=-0.17V$$\Delta G_1^0=-1\times F\times (-0.17V)$$C{u}^{+}+e\to Cu$$E_2^0=0.53V$$\Delta G^0=-1\times F\times (0.53V)=-0.53F$The $\Delta G^0$ for the reaction: $C{u}^{+}+I^{-}\to CuI$$\Delta G^0=\Delta G_2^0-\Delta G_1^0$= $-0.53F-(0.17F)=-0.70F$= $-0.70\times 96500J=-67550J$= $-67.55kJ$Type of Electrochemical cell:(i)Electrode cell(ii)Concentration cellElectrode cell:Different electrolytes are used in both half cell.Resultant cell emf determined by $E_{cell}=E_{cell}^o-\frac{0.0591}{n}\log Q$Concentration cell:Same type of electrolytes used in both half cell but concentration of these electrolytes are different.Resultant cell emf determined by $E_{cell}=E_{cell}^0-\frac{0.0591}{n}\log Q$Here,$\therefore E_{cell}=\frac{0.0591}{n}\log \frac{1}{Q}$Illustration 7.Stronger the oxidizing agent, greater is the(A) Standard reduction potential(B) Standard oxidation potential(C) Ionic nature(D)NoneSolution:(A)Illustration 8.E°_RP for Fe⁺² / Fe and Sn⁺² / Sn are – 0.44 and 0.14 volts respectively. The standard emf for cellFe⁺² + Sn $\to$ Sn⁺² + Fe is(A) + 0.30 V(B) 𠄰.58 V(C) + 0.58 V(D) 𠄰.30 VSolution:Fe⁺² + Sn $\to$ Sn⁺² + Fe$E_{cell}^0=E_{F{e}^{+2}/Fe}^0+E_{Sn/S{n}^{+2}}^0$= – 0.44 – 0.14 = – 0.58 VHence, (B) is correct.Illustration 9.EMF of the cellCd | CdSO₄ | H₂SO₄ (0.01 M) | H₂ (1 atm) | Ptis + 0.362 Volt at 25° C. The $E_{C{d}^{+2}/Cd}^o$ is 𠄰.403 Volt at same temperature. Calculate the solubility product of CdSO₄.Solution:Anode half cell: Cd $\to$ Cd⁺² (aq) + 2e^–Cathode half cell: 2H⁺ (aq) + 2e^– $\to$ H₂—————————————————————————— Net cell reaction:Cd + 2H⁺ (0.02 M) $\to$ H₂ (1 atm) + Cd⁺² (aq)$\begin{array}{l}\therefore \end{array}$E_cell =$E_{cell}^0-\frac{0.0591}{2}\log \frac{[C{d}^{+2}](P_{H_2})}{{[H^{+}]}^2}$

+ 0.362 = + 0.403 –$\frac{0.0591}{2}\log \frac{[C{d}^{+2}](1)}{{(0.02)}^2}$

–0.041 = –$\frac{0.0591}{2}\log \frac{[C{d}^{+2}]}{4\times {10}^{-4}}$

1.3875 = log $\frac{[C{d}^{+2}]}{4\times {10}^{-4}}$ $\Rightarrow$ $\frac{[C{d}^{+2}]}{4\times {10}^{-4}}$=24.41

$\therefore$[Cd⁺²] = 9.764 × 10^-3 M$\therefore$K_sp of Cd SO₄ = [Cd⁺²] $[S{O}_4^{-2}]$

= (9.764 × 10^–3) (0.01)

= 9.764 × 10^–5 M²