When acidulated water ($\mathrm{dil}. {\mathrm{H}}_2{\mathrm{SO}}_4$ solution) is electrolysed, will the $\mathrm{pH}$ of the solution be affected? Justify your answer.

(i) Electrochemical Cell: A device that converts chemical energy into electrical energy.

(ii) Anode: In electrochemical cells, the anode is the electrode at which oxidation takes place. It is the negative terminal.

(iii) Cathode: In electrochemical cells, the cathode is the electrode at which reduction takes place. It is the positive terminal.

2. Thermodynamic and electrochemical equations:

(i) EMF of the cell or ${\mathrm{E}}_{\mathrm{cell} }= {\mathrm{E}}_{\mathrm{cathode} }-{\mathrm{E}}_{\mathrm{anode} }$.

(ii) Thermodynamic Efficiency of Cell: It is the ratio of the Gibb’s energy change to the enthalpy change of the cell reaction $\mathrm{\eta } = \mathrm{\Delta G}/\mathrm{\Delta H}$.

(iii) Electrochemical Series: Arrangement of various elements and electrode reactions in the increasing order of their reduction potentials.

(iv) Nernst Equation:

${\mathrm{E}}_{{\mathrm{M}}^{\mathrm{n}+}/\mathrm{M}}={\mathrm{E}}^{\ominus }{\mathrm{M}}^{\mathrm{n}+/\mathrm{M}}+\frac{2.303\mathrm{RT}}{\mathrm{nF}}\log \frac{\left[{\mathrm{M}}^{\mathrm{n}+}\right]}{\left[\mathrm{M}\right]}$

                               or

${\mathrm{E}}_{{\mathrm{M}}^{\mathrm{n}+}/\mathrm{M}}={\mathrm{E}}^{\ominus }+\frac{0.059}{\mathrm{n}}\log \frac{\left[{\mathrm{M}}^{\mathrm{n}+}\right]}{\left[\mathrm{M}\right]}$

${{\mathrm{E}}^{\ominus }}_{\mathrm{cell}}=\frac{0.059}{\mathrm{n}}{\mathrm{logK}}_{\mathrm{c}}=\frac{2.303\mathrm{RT}}{\mathrm{nF}}{\mathrm{logK}}_{\mathrm{c}}$  (${\mathrm{K}}_{\mathrm{c}}$ is equiibrium constant for cell reaction)

(v) ∆rGo=-nFEcello

(vi) Fuel Cell: A device which converts chemical energy of a fuel directly into electrical energy.

3. Conduction in electrolytes:

(i) Electrolyte : A substance that dissociates in solution to produce ions and hence conducts electricity in dissolved state or molten state.

(ii) Conductivity (k): Inverse of resistivity. Conductivity of a material in S m–1 is its conductance when it is $1 \mathrm{m}$ long and its area of cross section is 1 m2. Its units are ohm-1 cm1 or $\mathrm{S}$ cm1 or Sm1.

(iii) Molar Conductivity: Conductance of a solution containing one mole of the electrolyte, placed between two parallel electrodes one cm apart.

${\mathrm{\Lambda }}_{\mathrm{m}}\mathrm{=}\frac{\mathrm{\kappa }}{\mathrm{C}}$ ${\mathrm{\Lambda }}_{\mathrm{m}}\left({\mathrm{Scm}}^2{\mathrm{mol}}^{-1}\right) =\frac{\mathrm{1000 }\left({\mathrm{cm}}^3/\mathrm{L}\right) \times \mathrm{\kappa }(\mathrm{S} {\mathrm{cm}}^{-\mathrm{1}})}{\mathrm{Molarity} \mathrm{(}\mathrm{mol}/\mathrm{L}\mathrm{)}}$

(iv) $\text{Conductivity}\left(\mathrm{\kappa }\right)\mathrm{=}\frac{ \text{Cell Constant} \left(\mathrm{G}*\right)}{ \text{Resistance} \left(\mathrm{R}\right)}=\text{Cell Constant}\left(\mathrm{G}*\right)\times \text{Conductance}\left(\mathrm{G}\right)$

(v) Molar conductivity of an electrolyte increases with dilution.

(vi) Limiting Molar Conductivity $\left({\mathrm{\Lambda }}_{\mathrm{m}}^{°}\right)$. It is the molar conductivity of electrolyte at infinite dilution.

(vii) Kohlrausch’s Law: The molar conductivity of an electrolyte at infinite dilution is equal to the sum of the ionic conductivities of the individual ions.

${\mathrm{\Lambda }}_{\mathrm{m}}^{\mathrm{°}} = {\mathrm{v}}_{\mathrm{+}}{\mathrm{\lambda }}_{\mathrm{+}}^{\mathrm{°}} + {\mathrm{v}}_{-}{\mathrm{\lambda }}_{-}^{\mathrm{°}}$

(viii) Degree of Dissociation $\left(\mathrm{\alpha }\right)$: Fraction of total number of molecules that dissociates in solution.

$\mathrm{\alpha }=\frac{{\mathrm{\Lambda }}_{\mathrm{m}}^{\mathrm{c}}}{{\mathrm{\Lambda }}_{\mathrm{m}}^{°}}$

(ix) Potential Gradient: It is the ratio of the potential applied across the electrodes to the distance between the electrodes.

(x) Electrolysis: The process of decomposition of electrolyte as a result of passage of electricity through its aqueous solution or through its molten state.