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 a $70 \%$ efficient hydrogen-oxygen fuel cell working under standard conditions at $1 \ bar$ and $298 \ K$. Its cell reaction is
$H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_2O(\ell)$
The work derived from the cell on the consumption of $1.0 \times 10^{-3} \ mol$ of $H_{2(g)}$ is used to compress $1.00 \ mol$ of a monoatomic ideal gas in a thermally insulated container. What is the change in the temperature (in $K$) of the ideal gas?
The standard reduction potentials for the two half-cells are given below.
$O_{2(g)} + 4H^{+}(aq.) + 4e^- \rightarrow 2H_2O(\ell), E^{\circ} = 1.23 \ V$
$2H^{+}(aq.) + 2e^- \rightarrow H_{2(g)}, E^{\circ} = 0.00 \ V$
Use $F = 96500 \ C \ mol^{-1}, R = 8.314 \ J \ mol^{-1} \ K^{-1}$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_2O(\ell)$
The work derived from the cell on the consumption of $1.0 \times 10^{-3} \ mol$ of $H_{2(g)}$ is used to compress $1.00 \ mol$ of a monoatomic ideal gas in a thermally insulated container. What is the change in the temperature (in $K$) of the ideal gas?
The standard reduction potentials for the two half-cells are given below.
$O_{2(g)} + 4H^{+}(aq.) + 4e^- \rightarrow 2H_2O(\ell), E^{\circ} = 1.23 \ V$
$2H^{+}(aq.) + 2e^- \rightarrow H_{2(g)}, E^{\circ} = 0.00 \ V$
Use $F = 96500 \ C \ mol^{-1}, R = 8.314 \ J \ mol^{-1} \ K^{-1}$
MediumIIT 2020
View SolutionFor a spontaneous reaction,determine the values of $\Delta G^o$,equilibrium constant $K$,and $E^o_{cell}$ respectively.
Medium
View SolutionMatch the following:
List-$I$ List-$II$ $(A)$ Potential of hydrogen electrode at $pH = 10$ $(I)$ $0.76 \ V$ $(B)$ $Cu^{2+}|Cu$ $(II)$ $0.059$ $(C)$ $Zn|Zn^{2+}$ $(III)$ $-0.591 \ V$ $(D)$ $\frac{2.303RT}{F}$ $(IV)$ $0.337 \ V$ $(V)$ $-0.76 \ V$
$(a)$ $A-III, B-I, C-II, D-V$
$(b)$ $A-II, B-V, C-I, D-IV$
$(c)$ $A-III, B-IV, C-I, D-II$
$(d)$ $A-V, B-I, C-IV, D-II$
| List-$I$ | List-$II$ |
|---|---|
| $(A)$ Potential of hydrogen electrode at $pH = 10$ | $(I)$ $0.76 \ V$ |
| $(B)$ $Cu^{2+}|Cu$ | $(II)$ $0.059$ |
| $(C)$ $Zn|Zn^{2+}$ | $(III)$ $-0.591 \ V$ |
| $(D)$ $\frac{2.303RT}{F}$ | $(IV)$ $0.337 \ V$ |
| $(V)$ $-0.76 \ V$ |
$(a)$ $A-III, B-I, C-II, D-V$
$(b)$ $A-II, B-V, C-I, D-IV$
$(c)$ $A-III, B-IV, C-I, D-II$
$(d)$ $A-V, B-I, C-IV, D-II$
Given $E_{Fe^{3+}/Fe^{2+}}^{\circ} = +0.76 \ V$ and $E_{I_{2}/I^{-}}^{\circ} = +0.55 \ V$. The equilibrium constant for the reaction taking place in a galvanic cell consisting of the above two electrodes is (Given $\frac{2.303 \ RT}{F} = 0.06 \ V$)
If $\Lambda^{0}_{NaCl} = 126 \ S \ cm^{2} \ mol^{-1}$,$\Lambda^{0}_{KBr} = 125 \ S \ cm^{2} \ mol^{-1}$,and $\Lambda^{0}_{KCl} = 150 \ S \ cm^{2} \ mol^{-1}$,then find $\Lambda^{0}_{NaBr}$ in $S \ cm^{2} \ mol^{-1}$.
Medium
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