For a cell involving one electron ${\mathrm{E}}_{\mathrm{cell}}^{⊝}=0.59\mathrm{V}$ at 298 K, the equilibrium constant for the cell reaction is: $\left[\mathrm{Given} \mathrm{that} \frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{V} \mathrm{at} \mathrm{T}=298 \mathrm{K}\right]$

The half cell reduction potential of a hydrogen electrode at pH = 10 will be :

For a reaction A(s) + 2B⁺ $\to$ A²⁺ + 2B(s) ;  K_C has been found to be 10¹². The ${\mathrm{E}}_{\mathrm{cell}}^{°}$ is : 

The Emf of the given cell is: 

Zn(s) | Zn⁺² (0.1M) || Sn⁺² (0.001M) | Sn(s)

Given ${\mathrm{E}}_{{\mathrm{Zn}}^{+2}/\mathrm{Zn}}^{\mathrm{o}}=-0.76 \mathrm{V}, {\mathrm{E}}_{{\mathrm{Sn}}^{2+}/\mathrm{Sn}}^{\mathrm{o}}=-0.14 \mathrm{V}$

The potential of hydrogen electrode having a pH = 10 is : 

Consider the following cell reaction 

2Fe(s) + ${\mathrm{O}}_2$(g) + 4${\mathrm{H}}^{+}$(aq) $\to$ 2${\mathrm{Fe}}^{2+}$(aq) + 2${\mathrm{H}}_2\mathrm{O}$(l)

E° = 1.67 V, At [${\mathrm{Fe}}^{2+}$] = 10${}^{-3}$ M, ${\mathrm{P}}_{{\mathrm{O}}_2}$ = 0.1 atm and pH = 3, the cell potential at 25°C is : 

Following limiting molar conductivities are given as

${\lambda }_m^0\left(H_{2}S{O}_4\right)=x Sc{m}^2 mo{l}^{-1}$

${\lambda }_m^0\left(K_{2}S{O}_4\right)=y Sc{m}^2 mo{l}^{-1}$

${\lambda }_m^0\left(C{H}_{3}COOK\right)=z Sc{m}^2 mo{l}^{-1}$

${\lambda }_m^0(in Sc{m}^2 mo{l}^{-1} ) for C{H}_{3}COOH$ will be-

$\mathrm{For} \mathrm{the} \mathrm{cell} \mathrm{Pt}\left(\mathrm{s}\right)\left|{\mathrm{Br}}_2\left(\mathrm{l}\right)\right|\left|{\mathrm{Br}}^{-}\left(0.010 \mathrm{M}\right)\right|\left|{\mathrm{H}}^{+}\left(0.030 \mathrm{M}\right)\right|\left|{\mathrm{H}}_2\left(\mathrm{g}\right)\left(1 \mathrm{bar}\right)\right|\mathrm{Pt}\left(\mathrm{s}\right)$,

If the concentration of ${\mathrm{Br}}^{-}$ becomes 2 times and the concentration of ${\mathrm{H}}^{+}$ becomes half of the initial value, then emf of the cell

The specific conductance (K) of 0.02 M aqueous acetic acid solution at 298 K is $1.65 \times {10}^{-4}$ S ${\mathrm{cm}}^{-1}$. The degree of dissociation of acetic acid is [Given: Equivalent conductance at infinite dilution of ${\mathrm{H}}^{+}$ = 349.1 S ${\mathrm{cm}}^2$ ${\mathrm{mol}}^{-1}$ and ${\mathrm{CH}}_3{\mathrm{COO}}^{-}$ = 40.9 S ${\mathrm{cm}}^2$ ${\mathrm{mol}}^{-1}$)

The equilibrium constant of a 2 electron redox reaction at 298 K is 3.8 x ${10}^{-3}$. The cell potential $\mathrm{E}°$ (in V) and the free energy change $∆$G$°$ (in kJ ${\mathrm{mol}}^{-1}$) for this equilibrium respectively, are -

The specific conductance of 0.01 M solution of the weak monobasic acid is 0.20 x ${10}^{-3}$ S ${\mathrm{cm}}^{-1}$. The dissociation constant of the acid is-

[Given  ${\Lambda }_{\mathrm{HA}}^{\infty }$ = 400 S ${\mathrm{cm}}^2$ ${\mathrm{mol}}^{-1}$]

Equivalent conductance of saturated ${\mathrm{BaSO}}_4$ solution is 400 ${\mathrm{ohm}}^{-1} {\mathrm{cm}}^{2 }{\mathrm{equivalent}}^{-1}$ and it's specific conductance is $8 \times {10}^{-5} {\mathrm{ohm}}^{-1} {\mathrm{cm}}^{-1}$; hence solubility product ${\mathrm{K}}_{\mathrm{sp}} \mathrm{of} {\mathrm{BaSO}}_4$ is : 

Aluminium oxide may be electrolysed at 1000 C to furnish aluminium metal (Atomic
mass = 27 amu; 1 Faraday = 96,500 Coulombs). The cathode reaction is : ${\mathrm{Al}}^{3+}+3{\mathrm{e}}^{-}\to \mathrm{Al}$

To prepare 5.12 kg of aluminium metal by this method would require : 

A fuel cell develops an electrical potential from the combustion of butane at 1 bar and 298 K 

${\mathrm{C}}_4{\mathrm{H}}_{10}\left(\mathrm{g}\right)+6.5{\mathrm{O}}_2\left(\mathrm{g}\right)\to 4{\mathrm{CO}}_2\left(\mathrm{g}\right)+5{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right);$ $\mathrm{E}°$ of a cell is: 

(Given $∆G^o =-2746kJ/mole$ )

A gas X at 1 atm is bubbled through a solution containing a mixture of 1M Y^- and 1M Z^- at 25 ^oC . If the reduction potential of Z>Y>X, then : 

The standard Emf of a galvanic cell involving cell reaction with n=2 is found to be 0.295 V at 25$°$C. The equilibrium constant of the reaction would be : 
(Given F=96500 C mol^-1; R=8.314 JK^-1 mol^-1)

$∆\mathrm{G}°$ for the reaction given below is: 

$\frac{1}{2}\underset{}{\mathrm{A}\left(\mathrm{g}\right) }+\frac{3}{2}\underset{}{\mathrm{B}\left(\mathrm{g}\right) }\rightleftharpoons \underset{}{\mathrm{C}\left(\mathrm{g}\right)}$

(${\mathrm{K}}_{\mathrm{eq}}=826 {\mathrm{atm}}^{-1} \mathrm{at} 298 \mathrm{K}$)

For the electrolysis of CuSO₄ solution the correct option is: 

Molten sodium chloride conducts electricity due to the presence of : 

One liter of 0.5 M KCl solution is electrolyzed for one minute in a current of 1.608 mA. Considering 100% efficiency, the pH of the resulting solution will be : 

The reaction from following having top position in EMF series (Std.red. potential) according to their electrode potential at 298 K is: 

Absolute electrode potential of an electrode can't be measured because :

 E⁰_cell for  feasible type cell reaction is : 

E⁰_cell for  non-feasible type cell reaction is :

 E_cell =0 or ${∆}_r$G =0  this condition is applicable on : 

The negative sign in the expression E⁰_Zn²⁺/Zn=— 0.76 V indicates : 

Electrode potential is potential difference between the -

How will the pH of brine (aq NaCl solution) be affected when it is electrolysed?

The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500 Ω. The cell constant if conductivity of 0.001M KCl solution at 298 K is 0.146 ×10^-3  Scm^-1 is: 

The conductivity of 0.00241 M acetic acid is 7.896 × 10^–5 S cm^–1. If ${\mathrm{\Lambda }}_{\mathrm{m}}^0$ for acetic acid is 390.5 S cm² mol^–1, the dissociation constant will be -

$$\begin{gathered} 1. 2.45 \times {10}^{-5 }\mathrm{mol} {\mathrm{L}}^{-1} \\ 2. 1.86 \times {10}^{-5} \mathrm{mol} {\mathrm{L}}^{-1} \\ 3. 3.72 \times {10}^{-4} \mathrm{mol} {\mathrm{L}}^{-1} \\ 4. 2.12 \times {10}^{-6} \mathrm{mol} {\mathrm{L}}^{-1} \end{gathered}$$