Electrochemistry

The pH and conductance of a weak acid (HX) were found to be 5 and $4\times10^{-5}$ S, respectively. The conductance was measured under standard condition using a cell where the electrode plates having a surface area of 1 cm$^2$ were at a distance of 15 cm apart. The value of the limiting molar conductivity is S m$^2$ mol$^{-1}$. (nearest integer)

(Given: degree of dissociation of the weak acid ($\alpha$) $\ll 1$)

MX is a sparingly soluble salt that follows the given solubility equilibrium at 298 K

$\ce{MX(s) <=> M+(aq) + X^-(aq)}$; $K_\mathrm{sp} = 10^{-10}$

If the standard reduction potential for

$\ce{M+(aq) ->[+e^-] M(s)}$ is $\left(E^\ominus_{\mathrm{M^+/M}}\right) = 0.79$ V, then the value of the standard reduction potential for the metal/metal insoluble salt electrode $E^\ominus_{\mathrm{X^-/MX(s)/M}}$ is mV. (nearest integer)

[Given: $\dfrac{2.303RT}{F} = 0.059$ V]

Consider the following electrochemical cell at 298K

$\mathrm{Pt}\,|\,\ce{HSnO2^-}(\mathrm{aq})\,|\,\ce{Sn(OH)6^{2-}}(\mathrm{aq})\,|\,\ce{Bi2O3}(\mathrm{s})\,|\,\mathrm{Bi}(\mathrm{s})$.

If the reaction quotient at a given time is $10^{6}$, then the cell EMF ($E_\mathrm{cell}$) is $\times\,10^{-1}$ V (Nearest integer).

Given the standard half-cell reduction potential as

$E^{0}_{\ce{Bi2O3}/\mathrm{Bi},\,\ce{OH-}} = -0.44$ V and

$E^{0}_{\ce{Sn(OH)6^{2-}}/\ce{HSnO2^-},\,\ce{OH-}} = -0.90$ V

Consider the following reduction processes :

$\ce{Al^3+ + 3e- -> Al(s)}$, $E^\circ = -1.66$ V

$\ce{Fe^3+ + e- -> Fe^2+}$, $E^\circ = +0.77$ V

$\ce{Co^3+ + e- -> Co^2+}$, $E^\circ = +1.81$ V

$\ce{Cr^3+ + 3e- -> Cr(s)}$, $E^\circ = -0.74$ V

The tendency to act as reducing agent decreases in the order :

Consider the following electrochemical cell :

Pt $\mid$ $\ce{O2}$(g) (1bar) $\mid$ HCl (aq) $\parallel$ $\ce{M^2+}$(aq, 1.0 M) $\mid$ M(s)

The pH above which, oxygen gas would start to evolve at anode is (nearest integer).

[Given : $E^{0}_{\mathrm{M^{2+}/M}} = 0.994$ V and $E^{0}_{\mathrm{O_2/H_2O}} = 1.23$ V (standard reduction potential) and $\dfrac{RT}{F}(2.303) = 0.059$ V at the given condition]

• (A)Concentration of $\ce{Fe^{2+}}$ is increased.
• (B)Concentration of $\ce{Fe^{3+}}$ is decreased
• (C)Concentration of $\ce{Fe^{2+}}$ is decreased
• (D)Concentration of $\ce{Fe^{3+}}$ is increased
• (E)Concentration of $\ce{Cl-}$ is increased

Electricity is passed through an acidic solution of $\ce{Cu^{2+}}$ till all the $\ce{Cu^{2+}}$ was exhausted, leading to the deposition of 300 mg of Cu metal. However, a current of 600 mA was continued to pass through the same solution for another 28 minutes by keeping the total volume of the solution fixed at 200 mL. The total volume of oxygen evolved at STP during the entire process is mL. (Nearest integer)

[Given :

$\ce{Cu^{2+}(aq) + 2e^- -> Cu(s)}$   $E^0_\mathrm{red} = +0.34$ V

$\ce{O2(g) + 4H^+ + 4e^- -> 2H2O}$   $E^0_\mathrm{red} = +1.23$ V

Molar mass of Cu = 63.54 g mol$^{-1}$

Molar mass of $\ce{O2}$ = 32 g mol$^{-1}$

Faraday Constant = 96500 C mol$^{-1}$

Molar volume at STP = 22.4 L ]

Molar conductivity of a weak acid HQ of concentration 0.18 M was found to be 1/30 of the molar conductivity of another weak acid HZ with concentration of 0.02 M. If $\lambda^{0}_{\mathrm{Q}^-}$ happened to be equal with $\lambda^{0}_{\mathrm{Z}^-}$, then the difference of the $\mathrm{p}K_\mathrm{a}$ values of the two weak acids ($\mathrm{p}K_\mathrm{a}(\mathrm{HQ})-\mathrm{p}K_\mathrm{a}(\mathrm{HZ})$) is (Nearest integer).

[Given : degree of dissociation ($\alpha$) $\ll 1$ for both weak acids, $\lambda^\circ$ : limiting molar conductivity of ions]

A volume of x mL of 5 M $\ce{NaHCO3}$ solution was mixed with 10 mL of 2 M $\ce{H2CO3}$ solution to make an electrolytic buffer. If the same buffer was used in the following electrochemical cell to record a cell potential of 235.3 mV, then the value of x $=$ mL (nearest integer).

$\ce{Sn(s) | Sn(OH)6^2-}$ (0.5 M) $|$ $\ce{HSnO2-}$ (0.05 M) $|$ $\ce{OH-}$ $|$ $\ce{Bi2O3(s) | Bi(s)}$

Consider upto one place of decimal for intermediate calculations $$\left[\begin{array}{ll} \text{Given :}\qquad & E^\circ_{\ce{Sn(OH)6^2-}|\ce{HSnO2-}} = -0.9\ \mathrm{V} \\[2pt] & E^\circ_{\ce{Bi2O3}|\ce{Bi}} = -0.44\ \mathrm{V} \\[2pt] & \mathrm{p}K_{\mathrm{a}(\ce{H2CO3})} = 6.11 \\[2pt] & \dfrac{2.303RT}{F} = 0.059\ \mathrm{V} \\[6pt] & \text{Antilog}\,(1.29) = 19.5 \end{array}\right]$$

An electrochemical cell is constructed using half cells in the direction of spontaneous change

$\ce{Fe(OH)2(s) + 2e- -> Fe(s) + 2OH-(aq)}$   $E^0 = -0.88$ V

and $\ce{AgBr(s) + e- -> Ag(s) + Br-(aq)}$   $E^0 = +0.07$ V

Which of the following option is correct ?

Consider the following two half-cell reactions along with the standard reduction potential given :

$\ce{CO2 + 6H+ + 6e- -> CH3OH + H2O}$    $E^\circ_\mathrm{red} = 0.02$ V

$\dfrac{1}{2}$ $\ce{O2 + 2H+ + 2e- -> H2O}$    $E^\circ_\mathrm{red} = 1.23$ V

The fuel cell was set up using the above two reactions such that the cell operates under the standard condition of 1 bar pressure and 298 K temperature. The fuel cell works with 80% efficiency. If the work derived from the cell using 1 mol of $\ce{CH3OH}$ is used to compress an ideal gas isothermally against a constant pressure of 1kPa, then the change in the volume of the gas, $\lvert\Delta V\rvert$ = m$^3$. (Nearest integer)

Given : $F = 96500$ C mol$^{-1}$

An electrochemical cell consists of the following two redox couples,

$\mathrm{M}^{x+}$(aq)/M(s) [$E^\ominus_\mathrm{red} = +0.15$ V] and

$\ce{Fe^3+}$(aq)/Fe(s) [$E^\ominus_\mathrm{red} = -0.036$ V]

The cell EMF ($E_\mathrm{cell}$) is recorded to be 0.2057 V.

If the reaction quotient of the electrochemical reaction is found to be $10^{-2}$, then the value of x is (Nearest integer)

[Given : M is a p-block metal and $\dfrac{2.303RT}{F} = 0.059$ V]

One half cell in a voltaic cell is constructed by dipping silver rod in $\ce{AgNO3}$ solution of unknown concentration, other half cell is Zn rod dipped in 1 molar solution of $\ce{ZnSO4}$.

A voltage of 1.60 V is measured at 298 K for this cell. What is the concentration of $\ce{Ag+}$ ions used in terms of $\log x$ ($x = [\ce{Ag+}]$) ?

$E^\ominus_{\mathrm{Zn^{2+}/Zn}} = -0.76$ V, $E^\ominus_{\mathrm{Ag^{+}/Ag}} = +0.80$ V,

$\dfrac{2.303RT}{F} = 0.059$ V

At 298 K, the molar conductivity of x% (w/w) MX solution (aqueous) is 123.5 S cm$^2$ mol$^{-1}$. The conductance of same solution is $1.9\times10^{-3}$ S. The value of x is $\times\,10^{-2}$.

(Given : cell constant $= 1.3$ cm$^{-1}$; molar mass of MX is 75 g mol$^{-1}$, density of aqueous solution of MX at 298 K is 1.0 g mL$^{-1}$)

$\ce{BaSO4}$ is sparingly soluble in water. If the conductivity of the saturated $\ce{BaSO4}$ solution is $x$ S cm$^{-1}$ then the solubility product of $\ce{BaSO4}$ can be given as.

(Here $\Lambda_\mathrm{m} = \Lambda^\circ_\mathrm{m}$)

For a general redox reaction

Anode   $\mathrm{Red_1 \rightarrow Ox_1^{\,n_1^+} + n_1e^-}$

Cathode   $\mathrm{Ox_2 + n_2e^- \rightarrow Red_2^{\,n_2^-}}$

Which of the following statement is incorrect ?

Given at 298 K : $E^{\ominus}_{\mathrm{Fe^{2+}/Fe}} = X$ Volt

 $E^{\ominus}_{\mathrm{Fe^{3+}/Fe}} = Y$ Volt

The $E^{\ominus}_{\mathrm{Fe^{3+}/Fe^{2+}}}$ in Volt at 298 K is given by :