275815 Given the equilibrium constant $\left(K_{c}\right)$ of the reaction:
$\mathrm{Cu}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})$
is $10 \times 10^{15}$, calculate the $E_{\text {cell }}^{o}$ of this reaction at $298 \mathrm{~K}$.
$\left[2.303 \frac{\mathrm{RT}}{\mathrm{F}} \text { at } 298 \mathrm{~K}=0.059 \mathrm{~V}\right]$

275817 Net cell reaction of $\mathrm{Pt}\left \vert\mathrm{H}_{2}(640 \mathrm{~mm})\right \vert \mathrm{HCl} \mid \mathrm{H}_{2}$ $(510 \mathrm{~mm})$ pt.

275818 The standard reduction potential $E^{0}$ for half reactions are
$\mathrm{Zn} \rightarrow \mathbf{Z n}^{2+}+2 \mathrm{e}^{-} \quad ; \mathrm{E}^{\mathbf{0}}=+\mathbf{0 . 7 6 \mathrm { V }}$
$\mathrm{Fe} \rightarrow \mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \quad ; \mathrm{E}^{\mathbf{0}}=+\mathbf{0 . 4 1 V}$
The EMF of the cell reaction
$\mathrm{Fe}^{2+}+\mathbf{Z n} \rightarrow \mathbf{Z n}^{2+}+\mathbf{F e}$ is

275820 In following cell reaction
$\mathrm{Mg}(\mathrm{s})+2 \mathrm{Ag}^{+}(0.001 \mathrm{M}) \rightarrow \mathrm{Mg}^{2+}(0.20 \mathrm{M})+2 \mathrm{Ag}(\mathrm{s})$
Calcuate $E_{\text {cell }}$ for the reason $\left[E^{0}=3.17\right]$

275815 Given the equilibrium constant $\left(K_{c}\right)$ of the reaction:
$\mathrm{Cu}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})$
is $10 \times 10^{15}$, calculate the $E_{\text {cell }}^{o}$ of this reaction at $298 \mathrm{~K}$.
$\left[2.303 \frac{\mathrm{RT}}{\mathrm{F}} \text { at } 298 \mathrm{~K}=0.059 \mathrm{~V}\right]$

275817 Net cell reaction of $\mathrm{Pt}\left \vert\mathrm{H}_{2}(640 \mathrm{~mm})\right \vert \mathrm{HCl} \mid \mathrm{H}_{2}$ $(510 \mathrm{~mm})$ pt.

275818 The standard reduction potential $E^{0}$ for half reactions are
$\mathrm{Zn} \rightarrow \mathbf{Z n}^{2+}+2 \mathrm{e}^{-} \quad ; \mathrm{E}^{\mathbf{0}}=+\mathbf{0 . 7 6 \mathrm { V }}$
$\mathrm{Fe} \rightarrow \mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \quad ; \mathrm{E}^{\mathbf{0}}=+\mathbf{0 . 4 1 V}$
The EMF of the cell reaction
$\mathrm{Fe}^{2+}+\mathbf{Z n} \rightarrow \mathbf{Z n}^{2+}+\mathbf{F e}$ is

275820 In following cell reaction
$\mathrm{Mg}(\mathrm{s})+2 \mathrm{Ag}^{+}(0.001 \mathrm{M}) \rightarrow \mathrm{Mg}^{2+}(0.20 \mathrm{M})+2 \mathrm{Ag}(\mathrm{s})$
Calcuate $E_{\text {cell }}$ for the reason $\left[E^{0}=3.17\right]$

275815 Given the equilibrium constant $\left(K_{c}\right)$ of the reaction:
$\mathrm{Cu}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})$
is $10 \times 10^{15}$, calculate the $E_{\text {cell }}^{o}$ of this reaction at $298 \mathrm{~K}$.
$\left[2.303 \frac{\mathrm{RT}}{\mathrm{F}} \text { at } 298 \mathrm{~K}=0.059 \mathrm{~V}\right]$

275817 Net cell reaction of $\mathrm{Pt}\left \vert\mathrm{H}_{2}(640 \mathrm{~mm})\right \vert \mathrm{HCl} \mid \mathrm{H}_{2}$ $(510 \mathrm{~mm})$ pt.

275818 The standard reduction potential $E^{0}$ for half reactions are
$\mathrm{Zn} \rightarrow \mathbf{Z n}^{2+}+2 \mathrm{e}^{-} \quad ; \mathrm{E}^{\mathbf{0}}=+\mathbf{0 . 7 6 \mathrm { V }}$
$\mathrm{Fe} \rightarrow \mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \quad ; \mathrm{E}^{\mathbf{0}}=+\mathbf{0 . 4 1 V}$
The EMF of the cell reaction
$\mathrm{Fe}^{2+}+\mathbf{Z n} \rightarrow \mathbf{Z n}^{2+}+\mathbf{F e}$ is

275820 In following cell reaction
$\mathrm{Mg}(\mathrm{s})+2 \mathrm{Ag}^{+}(0.001 \mathrm{M}) \rightarrow \mathrm{Mg}^{2+}(0.20 \mathrm{M})+2 \mathrm{Ag}(\mathrm{s})$
Calcuate $E_{\text {cell }}$ for the reason $\left[E^{0}=3.17\right]$

275815 Given the equilibrium constant $\left(K_{c}\right)$ of the reaction:
$\mathrm{Cu}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})$
is $10 \times 10^{15}$, calculate the $E_{\text {cell }}^{o}$ of this reaction at $298 \mathrm{~K}$.
$\left[2.303 \frac{\mathrm{RT}}{\mathrm{F}} \text { at } 298 \mathrm{~K}=0.059 \mathrm{~V}\right]$

275817 Net cell reaction of $\mathrm{Pt}\left \vert\mathrm{H}_{2}(640 \mathrm{~mm})\right \vert \mathrm{HCl} \mid \mathrm{H}_{2}$ $(510 \mathrm{~mm})$ pt.

275818 The standard reduction potential $E^{0}$ for half reactions are
$\mathrm{Zn} \rightarrow \mathbf{Z n}^{2+}+2 \mathrm{e}^{-} \quad ; \mathrm{E}^{\mathbf{0}}=+\mathbf{0 . 7 6 \mathrm { V }}$
$\mathrm{Fe} \rightarrow \mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \quad ; \mathrm{E}^{\mathbf{0}}=+\mathbf{0 . 4 1 V}$
The EMF of the cell reaction
$\mathrm{Fe}^{2+}+\mathbf{Z n} \rightarrow \mathbf{Z n}^{2+}+\mathbf{F e}$ is

275820 In following cell reaction
$\mathrm{Mg}(\mathrm{s})+2 \mathrm{Ag}^{+}(0.001 \mathrm{M}) \rightarrow \mathrm{Mg}^{2+}(0.20 \mathrm{M})+2 \mathrm{Ag}(\mathrm{s})$
Calcuate $E_{\text {cell }}$ for the reason $\left[E^{0}=3.17\right]$