02. Cell Constant
ELECTROCHEMISTRY

276096 The emf of the Daniel Cell $\mathrm{Zn}\left \vert\mathrm{ZnSO}_{4}(0.01 \mathrm{M})\right \vert\left \vert\mathrm{CuSO}_{4}(\mathbf{1 M})\right \vert \mathrm{Cu}$ at $298 \mathrm{~K}$ is $E_{1}$. When concentration of $\mathrm{ZnSO}_{4}$ is changed to $1 \mathrm{M}$ and that of $\mathrm{CuSO}_{4}$ is changed to $0.01 \mathrm{M}$. the emf changed to $E_{2}$. Then find the relationship between $E_{1}$ and $E_{2}$.

1 $\mathrm{E}_{1}>\mathrm{E}_{2}$
2 $\mathrm{E}_{1}<\mathrm{E}_{2}$
3 $\mathrm{E}_{1}=\mathrm{E}_{2}$
4 $\mathrm{E}_{2}=0 \neq \mathrm{E}_{1}$
ELECTROCHEMISTRY

276097 Calculate the maximum work that can be obtained from the cell,
$\mathbf{Z n}\left \vert\mathbf{Z n}^{2+}(\mathbf{1} \mathbf{M})\right \vert\left \vert\mathbf{A g}^{+}(\mathbf{1 M})\right \vert \mathbf{A g}$
Where $\mathrm{E}^{\circ}{ }_{\mathrm{Zn}^{2+} \mid \mathrm{Zn}}=-0.76 \mathrm{~V}$ and $\mathrm{E}^{\circ}{ }_{\mathrm{Ag}^{+} \mid \mathrm{Ag}}=0.80 \mathrm{~V}$

1 $-301.080 \mathrm{~kJ}$
2 $201.830 \mathrm{~kJ}$
3 $112.830 \mathrm{~kJ}$
4 $212.630 \mathrm{~kJ}$
ELECTROCHEMISTRY

276098 For a cell reaction involving a two electron change, the standard emf of the cell is found to be $0.295 \mathrm{~V}$ at $25^{\circ} \mathrm{C}$. The equilibrium constant of the reaction at $25^{\circ} \mathrm{C}$ will be

1 $1 \times 10^{-10}$
2 $29.5 \times 10^{-2}$
3 10
4 $1 \times 10^{10}$
ELECTROCHEMISTRY

276099 Given the data at $25^{\circ} \mathrm{C}$
$\mathrm{Ag}+\mathrm{I}^{-} \rightarrow \mathrm{AgI}+\mathrm{e}^{-} ; \mathrm{E}^{\mathbf{0}}=\mathbf{0 . 1 5 2} \mathrm{V}$
$\mathrm{Ag} \rightarrow \mathrm{Ag}^{+}+\mathrm{e}^{-} ; \mathrm{E}^{\mathrm{o}}=-0.8 \mathrm{~V}$
What is the value of $\log K_{\text {sp }}$ for AgI?
$\left(2.303 \frac{\mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{~V}\right)$

1 -8.12
2 +8.612
3 -37.83
4 -16.13
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ELECTROCHEMISTRY

276096 The emf of the Daniel Cell $\mathrm{Zn}\left \vert\mathrm{ZnSO}_{4}(0.01 \mathrm{M})\right \vert\left \vert\mathrm{CuSO}_{4}(\mathbf{1 M})\right \vert \mathrm{Cu}$ at $298 \mathrm{~K}$ is $E_{1}$. When concentration of $\mathrm{ZnSO}_{4}$ is changed to $1 \mathrm{M}$ and that of $\mathrm{CuSO}_{4}$ is changed to $0.01 \mathrm{M}$. the emf changed to $E_{2}$. Then find the relationship between $E_{1}$ and $E_{2}$.

1 $\mathrm{E}_{1}>\mathrm{E}_{2}$
2 $\mathrm{E}_{1}<\mathrm{E}_{2}$
3 $\mathrm{E}_{1}=\mathrm{E}_{2}$
4 $\mathrm{E}_{2}=0 \neq \mathrm{E}_{1}$
ELECTROCHEMISTRY

276097 Calculate the maximum work that can be obtained from the cell,
$\mathbf{Z n}\left \vert\mathbf{Z n}^{2+}(\mathbf{1} \mathbf{M})\right \vert\left \vert\mathbf{A g}^{+}(\mathbf{1 M})\right \vert \mathbf{A g}$
Where $\mathrm{E}^{\circ}{ }_{\mathrm{Zn}^{2+} \mid \mathrm{Zn}}=-0.76 \mathrm{~V}$ and $\mathrm{E}^{\circ}{ }_{\mathrm{Ag}^{+} \mid \mathrm{Ag}}=0.80 \mathrm{~V}$

1 $-301.080 \mathrm{~kJ}$
2 $201.830 \mathrm{~kJ}$
3 $112.830 \mathrm{~kJ}$
4 $212.630 \mathrm{~kJ}$
ELECTROCHEMISTRY

276098 For a cell reaction involving a two electron change, the standard emf of the cell is found to be $0.295 \mathrm{~V}$ at $25^{\circ} \mathrm{C}$. The equilibrium constant of the reaction at $25^{\circ} \mathrm{C}$ will be

1 $1 \times 10^{-10}$
2 $29.5 \times 10^{-2}$
3 10
4 $1 \times 10^{10}$
ELECTROCHEMISTRY

276099 Given the data at $25^{\circ} \mathrm{C}$
$\mathrm{Ag}+\mathrm{I}^{-} \rightarrow \mathrm{AgI}+\mathrm{e}^{-} ; \mathrm{E}^{\mathbf{0}}=\mathbf{0 . 1 5 2} \mathrm{V}$
$\mathrm{Ag} \rightarrow \mathrm{Ag}^{+}+\mathrm{e}^{-} ; \mathrm{E}^{\mathrm{o}}=-0.8 \mathrm{~V}$
What is the value of $\log K_{\text {sp }}$ for AgI?
$\left(2.303 \frac{\mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{~V}\right)$

1 -8.12
2 +8.612
3 -37.83
4 -16.13
ELECTROCHEMISTRY

276096 The emf of the Daniel Cell $\mathrm{Zn}\left \vert\mathrm{ZnSO}_{4}(0.01 \mathrm{M})\right \vert\left \vert\mathrm{CuSO}_{4}(\mathbf{1 M})\right \vert \mathrm{Cu}$ at $298 \mathrm{~K}$ is $E_{1}$. When concentration of $\mathrm{ZnSO}_{4}$ is changed to $1 \mathrm{M}$ and that of $\mathrm{CuSO}_{4}$ is changed to $0.01 \mathrm{M}$. the emf changed to $E_{2}$. Then find the relationship between $E_{1}$ and $E_{2}$.

1 $\mathrm{E}_{1}>\mathrm{E}_{2}$
2 $\mathrm{E}_{1}<\mathrm{E}_{2}$
3 $\mathrm{E}_{1}=\mathrm{E}_{2}$
4 $\mathrm{E}_{2}=0 \neq \mathrm{E}_{1}$
ELECTROCHEMISTRY

276097 Calculate the maximum work that can be obtained from the cell,
$\mathbf{Z n}\left \vert\mathbf{Z n}^{2+}(\mathbf{1} \mathbf{M})\right \vert\left \vert\mathbf{A g}^{+}(\mathbf{1 M})\right \vert \mathbf{A g}$
Where $\mathrm{E}^{\circ}{ }_{\mathrm{Zn}^{2+} \mid \mathrm{Zn}}=-0.76 \mathrm{~V}$ and $\mathrm{E}^{\circ}{ }_{\mathrm{Ag}^{+} \mid \mathrm{Ag}}=0.80 \mathrm{~V}$

1 $-301.080 \mathrm{~kJ}$
2 $201.830 \mathrm{~kJ}$
3 $112.830 \mathrm{~kJ}$
4 $212.630 \mathrm{~kJ}$
ELECTROCHEMISTRY

276098 For a cell reaction involving a two electron change, the standard emf of the cell is found to be $0.295 \mathrm{~V}$ at $25^{\circ} \mathrm{C}$. The equilibrium constant of the reaction at $25^{\circ} \mathrm{C}$ will be

1 $1 \times 10^{-10}$
2 $29.5 \times 10^{-2}$
3 10
4 $1 \times 10^{10}$
ELECTROCHEMISTRY

276099 Given the data at $25^{\circ} \mathrm{C}$
$\mathrm{Ag}+\mathrm{I}^{-} \rightarrow \mathrm{AgI}+\mathrm{e}^{-} ; \mathrm{E}^{\mathbf{0}}=\mathbf{0 . 1 5 2} \mathrm{V}$
$\mathrm{Ag} \rightarrow \mathrm{Ag}^{+}+\mathrm{e}^{-} ; \mathrm{E}^{\mathrm{o}}=-0.8 \mathrm{~V}$
What is the value of $\log K_{\text {sp }}$ for AgI?
$\left(2.303 \frac{\mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{~V}\right)$

1 -8.12
2 +8.612
3 -37.83
4 -16.13
ELECTROCHEMISTRY

276096 The emf of the Daniel Cell $\mathrm{Zn}\left \vert\mathrm{ZnSO}_{4}(0.01 \mathrm{M})\right \vert\left \vert\mathrm{CuSO}_{4}(\mathbf{1 M})\right \vert \mathrm{Cu}$ at $298 \mathrm{~K}$ is $E_{1}$. When concentration of $\mathrm{ZnSO}_{4}$ is changed to $1 \mathrm{M}$ and that of $\mathrm{CuSO}_{4}$ is changed to $0.01 \mathrm{M}$. the emf changed to $E_{2}$. Then find the relationship between $E_{1}$ and $E_{2}$.

1 $\mathrm{E}_{1}>\mathrm{E}_{2}$
2 $\mathrm{E}_{1}<\mathrm{E}_{2}$
3 $\mathrm{E}_{1}=\mathrm{E}_{2}$
4 $\mathrm{E}_{2}=0 \neq \mathrm{E}_{1}$
ELECTROCHEMISTRY

276097 Calculate the maximum work that can be obtained from the cell,
$\mathbf{Z n}\left \vert\mathbf{Z n}^{2+}(\mathbf{1} \mathbf{M})\right \vert\left \vert\mathbf{A g}^{+}(\mathbf{1 M})\right \vert \mathbf{A g}$
Where $\mathrm{E}^{\circ}{ }_{\mathrm{Zn}^{2+} \mid \mathrm{Zn}}=-0.76 \mathrm{~V}$ and $\mathrm{E}^{\circ}{ }_{\mathrm{Ag}^{+} \mid \mathrm{Ag}}=0.80 \mathrm{~V}$

1 $-301.080 \mathrm{~kJ}$
2 $201.830 \mathrm{~kJ}$
3 $112.830 \mathrm{~kJ}$
4 $212.630 \mathrm{~kJ}$
ELECTROCHEMISTRY

276098 For a cell reaction involving a two electron change, the standard emf of the cell is found to be $0.295 \mathrm{~V}$ at $25^{\circ} \mathrm{C}$. The equilibrium constant of the reaction at $25^{\circ} \mathrm{C}$ will be

1 $1 \times 10^{-10}$
2 $29.5 \times 10^{-2}$
3 10
4 $1 \times 10^{10}$
ELECTROCHEMISTRY

276099 Given the data at $25^{\circ} \mathrm{C}$
$\mathrm{Ag}+\mathrm{I}^{-} \rightarrow \mathrm{AgI}+\mathrm{e}^{-} ; \mathrm{E}^{\mathbf{0}}=\mathbf{0 . 1 5 2} \mathrm{V}$
$\mathrm{Ag} \rightarrow \mathrm{Ag}^{+}+\mathrm{e}^{-} ; \mathrm{E}^{\mathrm{o}}=-0.8 \mathrm{~V}$
What is the value of $\log K_{\text {sp }}$ for AgI?
$\left(2.303 \frac{\mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{~V}\right)$

1 -8.12
2 +8.612
3 -37.83
4 -16.13