03. Nernst Equation
ELECTROCHEMISTRY

276122 At $298 \mathrm{~K}$, the standard electrode potentials of $\mathrm{Cu}^{2+} / \mathrm{Cu}, \mathrm{Zn}^{2+} / \mathrm{Zn}, \mathrm{Fe}^{2+} / \mathrm{Fe}$ and $\mathrm{Ag}^{+} / \mathrm{Ag}$ are 0.34 $\mathrm{V},-\mathbf{0 . 7 6} \mathrm{V},-\mathbf{0 . 4 4} \mathrm{V}$ and $0.80 \mathrm{~V}$ respectively.
On the basis of standard electrode potential, predict which of the following reaction can not occur?

1 $2 \mathrm{CuSO}_{4}$ (aq) $+2 \mathrm{Ag}$ (s) $\rightarrow 2 \mathrm{Cu}$ (s) $+\mathrm{Ag}_{2} \mathrm{SO}_{4}$ (aq)
2 $\mathrm{CuSO}_{4}(\mathrm{aq})+\mathrm{Zn}(\mathrm{s}) \mathrm{ZnSO}_{4}$ (aq) $+\mathrm{Cu}(\mathrm{s})$
3 $\mathrm{CuSO}_{4}$ (aq) $+\mathrm{Fe}(\mathrm{s}) \rightarrow \mathrm{FeSO}_{4}$ (aq) $+\mathrm{Cu}(\mathrm{s})$
4 $\mathrm{FeSO}_{4}$ (aq) $+\mathrm{Zn}(\mathrm{s}) \rightarrow \mathrm{ZnSO}_{4}(\mathrm{aq})+\mathrm{Fe}(\mathrm{s})$
ELECTROCHEMISTRY

276123 For
$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \stackrel{\text { yields }}{\longrightarrow} 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \text { O.E }$
$=1.33 \mathrm{Vat}\left[\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\right]=4.5$ millimole. $\left[\mathrm{Cr}^{3+}\right]=$ 1.5 millimole and $E=1.067 \mathrm{~V}$. Then calculate the $\mathrm{pH}$ of the solution.

1 2
2 3
3 2.5
4 1.5
ELECTROCHEMISTRY

276124 $\mathrm{H}_{2}(\mathrm{~g})+2 \mathrm{AgCl}(\mathrm{g}) \rightleftharpoons 2 \mathrm{Ag}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}), \mathrm{E}_{\text {cell }}^{\circ}$ at $25^{\circ} \mathrm{C}$ for the cell is $0.22 \mathrm{~V}$. The equilibrium constant at $25^{\circ} \mathrm{C}$ is

1 $2.8 \times 10^{7}$
2 $5.2 \times 10^{8}$
3 $2.8 \times 10^{5}$
4 $5.2 \times 10^{4}$
ELECTROCHEMISTRY

276125 For the reaction, $2 \mathrm{NH}_{3}(\mathrm{~s})+\mathrm{CO}_{2}(\mathrm{~g}) \rightleftharpoons$ $\mathrm{NH}_{2} \mathrm{CONH}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)$ find the value of equilibrium constant at $295 \mathrm{~K}$. Given, standard Gibbs energy change at the given temperature is $13.9 \mathrm{~kJ} \mathrm{~mol}^{-1}$.

1 $2.88 \times 10^{2}$
2 $2.58 \times 10^{2}$
3 $2.40 \times 10^{2}$
4 $2.65 \times 10^{2}$
ELECTROCHEMISTRY

276127 Standard cell voltage for the cell $\mathrm{Pb}\left \vert\mathrm{Pb}^{2+} \ \vert \mathrm{Sn}^{2+}\right \vert \mathrm{Sn}$ is $-0.01 \mathrm{~V}$. If the cell is to exhibit $E_{\text {cell }}=0$, the value of $\left[\mathrm{Sn}^{2+}\right] /\left[\mathrm{Pb}^{2+}\right]$ should be antilog of -

1 +0.3
2 0.5
3 1.5
4 -0.5
ELECTROCHEMISTRY

276122 At $298 \mathrm{~K}$, the standard electrode potentials of $\mathrm{Cu}^{2+} / \mathrm{Cu}, \mathrm{Zn}^{2+} / \mathrm{Zn}, \mathrm{Fe}^{2+} / \mathrm{Fe}$ and $\mathrm{Ag}^{+} / \mathrm{Ag}$ are 0.34 $\mathrm{V},-\mathbf{0 . 7 6} \mathrm{V},-\mathbf{0 . 4 4} \mathrm{V}$ and $0.80 \mathrm{~V}$ respectively.
On the basis of standard electrode potential, predict which of the following reaction can not occur?

1 $2 \mathrm{CuSO}_{4}$ (aq) $+2 \mathrm{Ag}$ (s) $\rightarrow 2 \mathrm{Cu}$ (s) $+\mathrm{Ag}_{2} \mathrm{SO}_{4}$ (aq)
2 $\mathrm{CuSO}_{4}(\mathrm{aq})+\mathrm{Zn}(\mathrm{s}) \mathrm{ZnSO}_{4}$ (aq) $+\mathrm{Cu}(\mathrm{s})$
3 $\mathrm{CuSO}_{4}$ (aq) $+\mathrm{Fe}(\mathrm{s}) \rightarrow \mathrm{FeSO}_{4}$ (aq) $+\mathrm{Cu}(\mathrm{s})$
4 $\mathrm{FeSO}_{4}$ (aq) $+\mathrm{Zn}(\mathrm{s}) \rightarrow \mathrm{ZnSO}_{4}(\mathrm{aq})+\mathrm{Fe}(\mathrm{s})$
ELECTROCHEMISTRY

276123 For
$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \stackrel{\text { yields }}{\longrightarrow} 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \text { O.E }$
$=1.33 \mathrm{Vat}\left[\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\right]=4.5$ millimole. $\left[\mathrm{Cr}^{3+}\right]=$ 1.5 millimole and $E=1.067 \mathrm{~V}$. Then calculate the $\mathrm{pH}$ of the solution.

1 2
2 3
3 2.5
4 1.5
ELECTROCHEMISTRY

276124 $\mathrm{H}_{2}(\mathrm{~g})+2 \mathrm{AgCl}(\mathrm{g}) \rightleftharpoons 2 \mathrm{Ag}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}), \mathrm{E}_{\text {cell }}^{\circ}$ at $25^{\circ} \mathrm{C}$ for the cell is $0.22 \mathrm{~V}$. The equilibrium constant at $25^{\circ} \mathrm{C}$ is

1 $2.8 \times 10^{7}$
2 $5.2 \times 10^{8}$
3 $2.8 \times 10^{5}$
4 $5.2 \times 10^{4}$
ELECTROCHEMISTRY

276125 For the reaction, $2 \mathrm{NH}_{3}(\mathrm{~s})+\mathrm{CO}_{2}(\mathrm{~g}) \rightleftharpoons$ $\mathrm{NH}_{2} \mathrm{CONH}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)$ find the value of equilibrium constant at $295 \mathrm{~K}$. Given, standard Gibbs energy change at the given temperature is $13.9 \mathrm{~kJ} \mathrm{~mol}^{-1}$.

1 $2.88 \times 10^{2}$
2 $2.58 \times 10^{2}$
3 $2.40 \times 10^{2}$
4 $2.65 \times 10^{2}$
ELECTROCHEMISTRY

276127 Standard cell voltage for the cell $\mathrm{Pb}\left \vert\mathrm{Pb}^{2+} \ \vert \mathrm{Sn}^{2+}\right \vert \mathrm{Sn}$ is $-0.01 \mathrm{~V}$. If the cell is to exhibit $E_{\text {cell }}=0$, the value of $\left[\mathrm{Sn}^{2+}\right] /\left[\mathrm{Pb}^{2+}\right]$ should be antilog of -

1 +0.3
2 0.5
3 1.5
4 -0.5
ELECTROCHEMISTRY

276122 At $298 \mathrm{~K}$, the standard electrode potentials of $\mathrm{Cu}^{2+} / \mathrm{Cu}, \mathrm{Zn}^{2+} / \mathrm{Zn}, \mathrm{Fe}^{2+} / \mathrm{Fe}$ and $\mathrm{Ag}^{+} / \mathrm{Ag}$ are 0.34 $\mathrm{V},-\mathbf{0 . 7 6} \mathrm{V},-\mathbf{0 . 4 4} \mathrm{V}$ and $0.80 \mathrm{~V}$ respectively.
On the basis of standard electrode potential, predict which of the following reaction can not occur?

1 $2 \mathrm{CuSO}_{4}$ (aq) $+2 \mathrm{Ag}$ (s) $\rightarrow 2 \mathrm{Cu}$ (s) $+\mathrm{Ag}_{2} \mathrm{SO}_{4}$ (aq)
2 $\mathrm{CuSO}_{4}(\mathrm{aq})+\mathrm{Zn}(\mathrm{s}) \mathrm{ZnSO}_{4}$ (aq) $+\mathrm{Cu}(\mathrm{s})$
3 $\mathrm{CuSO}_{4}$ (aq) $+\mathrm{Fe}(\mathrm{s}) \rightarrow \mathrm{FeSO}_{4}$ (aq) $+\mathrm{Cu}(\mathrm{s})$
4 $\mathrm{FeSO}_{4}$ (aq) $+\mathrm{Zn}(\mathrm{s}) \rightarrow \mathrm{ZnSO}_{4}(\mathrm{aq})+\mathrm{Fe}(\mathrm{s})$
ELECTROCHEMISTRY

276123 For
$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \stackrel{\text { yields }}{\longrightarrow} 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \text { O.E }$
$=1.33 \mathrm{Vat}\left[\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\right]=4.5$ millimole. $\left[\mathrm{Cr}^{3+}\right]=$ 1.5 millimole and $E=1.067 \mathrm{~V}$. Then calculate the $\mathrm{pH}$ of the solution.

1 2
2 3
3 2.5
4 1.5
ELECTROCHEMISTRY

276124 $\mathrm{H}_{2}(\mathrm{~g})+2 \mathrm{AgCl}(\mathrm{g}) \rightleftharpoons 2 \mathrm{Ag}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}), \mathrm{E}_{\text {cell }}^{\circ}$ at $25^{\circ} \mathrm{C}$ for the cell is $0.22 \mathrm{~V}$. The equilibrium constant at $25^{\circ} \mathrm{C}$ is

1 $2.8 \times 10^{7}$
2 $5.2 \times 10^{8}$
3 $2.8 \times 10^{5}$
4 $5.2 \times 10^{4}$
ELECTROCHEMISTRY

276125 For the reaction, $2 \mathrm{NH}_{3}(\mathrm{~s})+\mathrm{CO}_{2}(\mathrm{~g}) \rightleftharpoons$ $\mathrm{NH}_{2} \mathrm{CONH}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)$ find the value of equilibrium constant at $295 \mathrm{~K}$. Given, standard Gibbs energy change at the given temperature is $13.9 \mathrm{~kJ} \mathrm{~mol}^{-1}$.

1 $2.88 \times 10^{2}$
2 $2.58 \times 10^{2}$
3 $2.40 \times 10^{2}$
4 $2.65 \times 10^{2}$
ELECTROCHEMISTRY

276127 Standard cell voltage for the cell $\mathrm{Pb}\left \vert\mathrm{Pb}^{2+} \ \vert \mathrm{Sn}^{2+}\right \vert \mathrm{Sn}$ is $-0.01 \mathrm{~V}$. If the cell is to exhibit $E_{\text {cell }}=0$, the value of $\left[\mathrm{Sn}^{2+}\right] /\left[\mathrm{Pb}^{2+}\right]$ should be antilog of -

1 +0.3
2 0.5
3 1.5
4 -0.5
ELECTROCHEMISTRY

276122 At $298 \mathrm{~K}$, the standard electrode potentials of $\mathrm{Cu}^{2+} / \mathrm{Cu}, \mathrm{Zn}^{2+} / \mathrm{Zn}, \mathrm{Fe}^{2+} / \mathrm{Fe}$ and $\mathrm{Ag}^{+} / \mathrm{Ag}$ are 0.34 $\mathrm{V},-\mathbf{0 . 7 6} \mathrm{V},-\mathbf{0 . 4 4} \mathrm{V}$ and $0.80 \mathrm{~V}$ respectively.
On the basis of standard electrode potential, predict which of the following reaction can not occur?

1 $2 \mathrm{CuSO}_{4}$ (aq) $+2 \mathrm{Ag}$ (s) $\rightarrow 2 \mathrm{Cu}$ (s) $+\mathrm{Ag}_{2} \mathrm{SO}_{4}$ (aq)
2 $\mathrm{CuSO}_{4}(\mathrm{aq})+\mathrm{Zn}(\mathrm{s}) \mathrm{ZnSO}_{4}$ (aq) $+\mathrm{Cu}(\mathrm{s})$
3 $\mathrm{CuSO}_{4}$ (aq) $+\mathrm{Fe}(\mathrm{s}) \rightarrow \mathrm{FeSO}_{4}$ (aq) $+\mathrm{Cu}(\mathrm{s})$
4 $\mathrm{FeSO}_{4}$ (aq) $+\mathrm{Zn}(\mathrm{s}) \rightarrow \mathrm{ZnSO}_{4}(\mathrm{aq})+\mathrm{Fe}(\mathrm{s})$
ELECTROCHEMISTRY

276123 For
$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \stackrel{\text { yields }}{\longrightarrow} 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \text { O.E }$
$=1.33 \mathrm{Vat}\left[\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\right]=4.5$ millimole. $\left[\mathrm{Cr}^{3+}\right]=$ 1.5 millimole and $E=1.067 \mathrm{~V}$. Then calculate the $\mathrm{pH}$ of the solution.

1 2
2 3
3 2.5
4 1.5
ELECTROCHEMISTRY

276124 $\mathrm{H}_{2}(\mathrm{~g})+2 \mathrm{AgCl}(\mathrm{g}) \rightleftharpoons 2 \mathrm{Ag}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}), \mathrm{E}_{\text {cell }}^{\circ}$ at $25^{\circ} \mathrm{C}$ for the cell is $0.22 \mathrm{~V}$. The equilibrium constant at $25^{\circ} \mathrm{C}$ is

1 $2.8 \times 10^{7}$
2 $5.2 \times 10^{8}$
3 $2.8 \times 10^{5}$
4 $5.2 \times 10^{4}$
ELECTROCHEMISTRY

276125 For the reaction, $2 \mathrm{NH}_{3}(\mathrm{~s})+\mathrm{CO}_{2}(\mathrm{~g}) \rightleftharpoons$ $\mathrm{NH}_{2} \mathrm{CONH}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)$ find the value of equilibrium constant at $295 \mathrm{~K}$. Given, standard Gibbs energy change at the given temperature is $13.9 \mathrm{~kJ} \mathrm{~mol}^{-1}$.

1 $2.88 \times 10^{2}$
2 $2.58 \times 10^{2}$
3 $2.40 \times 10^{2}$
4 $2.65 \times 10^{2}$
ELECTROCHEMISTRY

276127 Standard cell voltage for the cell $\mathrm{Pb}\left \vert\mathrm{Pb}^{2+} \ \vert \mathrm{Sn}^{2+}\right \vert \mathrm{Sn}$ is $-0.01 \mathrm{~V}$. If the cell is to exhibit $E_{\text {cell }}=0$, the value of $\left[\mathrm{Sn}^{2+}\right] /\left[\mathrm{Pb}^{2+}\right]$ should be antilog of -

1 +0.3
2 0.5
3 1.5
4 -0.5
ELECTROCHEMISTRY

276122 At $298 \mathrm{~K}$, the standard electrode potentials of $\mathrm{Cu}^{2+} / \mathrm{Cu}, \mathrm{Zn}^{2+} / \mathrm{Zn}, \mathrm{Fe}^{2+} / \mathrm{Fe}$ and $\mathrm{Ag}^{+} / \mathrm{Ag}$ are 0.34 $\mathrm{V},-\mathbf{0 . 7 6} \mathrm{V},-\mathbf{0 . 4 4} \mathrm{V}$ and $0.80 \mathrm{~V}$ respectively.
On the basis of standard electrode potential, predict which of the following reaction can not occur?

1 $2 \mathrm{CuSO}_{4}$ (aq) $+2 \mathrm{Ag}$ (s) $\rightarrow 2 \mathrm{Cu}$ (s) $+\mathrm{Ag}_{2} \mathrm{SO}_{4}$ (aq)
2 $\mathrm{CuSO}_{4}(\mathrm{aq})+\mathrm{Zn}(\mathrm{s}) \mathrm{ZnSO}_{4}$ (aq) $+\mathrm{Cu}(\mathrm{s})$
3 $\mathrm{CuSO}_{4}$ (aq) $+\mathrm{Fe}(\mathrm{s}) \rightarrow \mathrm{FeSO}_{4}$ (aq) $+\mathrm{Cu}(\mathrm{s})$
4 $\mathrm{FeSO}_{4}$ (aq) $+\mathrm{Zn}(\mathrm{s}) \rightarrow \mathrm{ZnSO}_{4}(\mathrm{aq})+\mathrm{Fe}(\mathrm{s})$
ELECTROCHEMISTRY

276123 For
$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \stackrel{\text { yields }}{\longrightarrow} 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \text { O.E }$
$=1.33 \mathrm{Vat}\left[\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\right]=4.5$ millimole. $\left[\mathrm{Cr}^{3+}\right]=$ 1.5 millimole and $E=1.067 \mathrm{~V}$. Then calculate the $\mathrm{pH}$ of the solution.

1 2
2 3
3 2.5
4 1.5
ELECTROCHEMISTRY

276124 $\mathrm{H}_{2}(\mathrm{~g})+2 \mathrm{AgCl}(\mathrm{g}) \rightleftharpoons 2 \mathrm{Ag}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}), \mathrm{E}_{\text {cell }}^{\circ}$ at $25^{\circ} \mathrm{C}$ for the cell is $0.22 \mathrm{~V}$. The equilibrium constant at $25^{\circ} \mathrm{C}$ is

1 $2.8 \times 10^{7}$
2 $5.2 \times 10^{8}$
3 $2.8 \times 10^{5}$
4 $5.2 \times 10^{4}$
ELECTROCHEMISTRY

276125 For the reaction, $2 \mathrm{NH}_{3}(\mathrm{~s})+\mathrm{CO}_{2}(\mathrm{~g}) \rightleftharpoons$ $\mathrm{NH}_{2} \mathrm{CONH}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)$ find the value of equilibrium constant at $295 \mathrm{~K}$. Given, standard Gibbs energy change at the given temperature is $13.9 \mathrm{~kJ} \mathrm{~mol}^{-1}$.

1 $2.88 \times 10^{2}$
2 $2.58 \times 10^{2}$
3 $2.40 \times 10^{2}$
4 $2.65 \times 10^{2}$
ELECTROCHEMISTRY

276127 Standard cell voltage for the cell $\mathrm{Pb}\left \vert\mathrm{Pb}^{2+} \ \vert \mathrm{Sn}^{2+}\right \vert \mathrm{Sn}$ is $-0.01 \mathrm{~V}$. If the cell is to exhibit $E_{\text {cell }}=0$, the value of $\left[\mathrm{Sn}^{2+}\right] /\left[\mathrm{Pb}^{2+}\right]$ should be antilog of -

1 +0.3
2 0.5
3 1.5
4 -0.5