The electrochemical cell shown below is a concentration cell.
$M \mid M^{2+} (\text{saturated solution of a sparingly soluble salt, } MX_2) \mid M^{2+} (0.001 \ mol \ dm^{-3}) \mid M$
The emf of the cell depends on the difference in concentration of $M^{2+}$ ions at the two electrodes. The emf of the cell at $298 \ K$ is $0.059 \ V$.
$1.$ The solubility product $(K_{sp}; \ mol^3 \ dm^{-9})$ of $MX_2$ at $298 \ K$ based on the information available for the given concentration cell is (take $2.303 \times R \times 298 / F = 0.059 \ V$):
$(A) \ 1 \times 10^{-15} \quad (B) \ 4 \times 10^{-15}$
$(C) \ 1 \times 10^{-12} \quad (D) \ 4 \times 10^{-12}$
$2.$ The value of $\Delta G \ (kJ \ mol^{-1})$ for the given cell is (take $1 \ F = 96500 \ C \ mol^{-1}$):
$(A) \ -5.7 \quad (B) \ 5.7 \quad (C) \ 11.4 \quad (D) \ -11.4$
Give the answer for question $1$ and $2$.
$M \mid M^{2+} (\text{saturated solution of a sparingly soluble salt, } MX_2) \mid M^{2+} (0.001 \ mol \ dm^{-3}) \mid M$
The emf of the cell depends on the difference in concentration of $M^{2+}$ ions at the two electrodes. The emf of the cell at $298 \ K$ is $0.059 \ V$.
$1.$ The solubility product $(K_{sp}; \ mol^3 \ dm^{-9})$ of $MX_2$ at $298 \ K$ based on the information available for the given concentration cell is (take $2.303 \times R \times 298 / F = 0.059 \ V$):
$(A) \ 1 \times 10^{-15} \quad (B) \ 4 \times 10^{-15}$
$(C) \ 1 \times 10^{-12} \quad (D) \ 4 \times 10^{-12}$
$2.$ The value of $\Delta G \ (kJ \ mol^{-1})$ for the given cell is (take $1 \ F = 96500 \ C \ mol^{-1}$):
$(A) \ -5.7 \quad (B) \ 5.7 \quad (C) \ 11.4 \quad (D) \ -11.4$
Give the answer for question $1$ and $2$.
- A$(B, D)$
- B$(B, C)$
- C$(A, D)$
- D$(C, D)$
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Similar Questions
Consider the following data.
Electrolyte $\Lambda_m^\circ$ $(S\text{ cm}^2\text{ mol}^{-1})$ $BaCl_2$ $x_1$ $H_2SO_4$ $x_2$ $HCl$ $x_3$
$BaSO_4$ is sparingly soluble in water. If the conductivity of the saturated $BaSO_4$ solution is $x\text{ S cm}^{-1}$, then the solubility product of $BaSO_4$ can be given as (Here $\Lambda_m = \Lambda_m^\circ$)
| Electrolyte | $\Lambda_m^\circ$ $(S\text{ cm}^2\text{ mol}^{-1})$ |
|---|---|
| $BaCl_2$ | $x_1$ |
| $H_2SO_4$ | $x_2$ |
| $HCl$ | $x_3$ |
$BaSO_4$ is sparingly soluble in water. If the conductivity of the saturated $BaSO_4$ solution is $x\text{ S cm}^{-1}$, then the solubility product of $BaSO_4$ can be given as (Here $\Lambda_m = \Lambda_m^\circ$)
DifficultJEE MAIN 2026
View SolutionAt $300 \ K$,the $E_{cell}^{\circ}$ of $A_{(s)} + B^{2+}_{(aq)} \rightleftharpoons A^{2+}_{(aq)} + B_{(s)}$ is $1.0 \ V$. If $\Delta_r S^{\circ}$ of this reaction is $100 \ J \ K^{-1} \ mol^{-1}$,what is $\Delta_r H^{\circ}$ (in $kJ \ mol^{-1}$) of this reaction? $(F = 96500 \ C \ mol^{-1})$
Calculate the $K_C$ and $\Delta G^o$ for the chemical reaction :
$Ni_{(s)} + 2Ag_{(aq)}^{+} \to Ni_{(aq)}^{2+} + 2Ag_{(s)}$ $ [E^o = 1.05 \, V] $
$Ni_{(s)} + 2Ag_{(aq)}^{+} \to Ni_{(aq)}^{2+} + 2Ag_{(s)}$ $ [E^o = 1.05 \, V] $
Difficult
View SolutionWhich chemical substances are produced using electrochemistry?
Easy
View Solution
Column $I$
Column $II$
$(A)$. Kohlrausch law can calculate
$(P)$. $\frac{\Lambda_m^c}{\Lambda_m^o}$
$(B)$. Molar conductance $\Lambda_m$
$(Q)$. $\frac{1}{R} \times \frac{l}{A}$
$(C)$. Specific conductance $\kappa$
$(R)$. $\Lambda_m^o$ of $Ca_3(PO_4)_2$
$(D)$. Degree of ionization of weak electrolyte
$(S)$. $\frac{\kappa \times 1000}{M}$
Which of the following options shows the correct matches?
| Column $I$ | Column $II$ |
|---|---|
| $(A)$. Kohlrausch law can calculate | $(P)$. $\frac{\Lambda_m^c}{\Lambda_m^o}$ |
| $(B)$. Molar conductance $\Lambda_m$ | $(Q)$. $\frac{1}{R} \times \frac{l}{A}$ |
| $(C)$. Specific conductance $\kappa$ | $(R)$. $\Lambda_m^o$ of $Ca_3(PO_4)_2$ |
| $(D)$. Degree of ionization of weak electrolyte | $(S)$. $\frac{\kappa \times 1000}{M}$ |
Medium
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