Nernst equation and ECS Questions in English

(B) The cell reaction is: $Zn(s) + Fe^{2+}(aq) \rightarrow Zn^{2+}(aq) + Fe(s)$.
Here,$n = 2$.
Using the Nernst equation: $E_{cell} = E^0_{cell} - \frac{0.0591}{n} \log \frac{[Zn^{2+}]}{[Fe^{2+}]}$.
$0.2905 = E^0_{cell} - \frac{0.0591}{2} \log \frac{0.01}{0.001}$.
$0.2905 = E^0_{cell} - 0.02955 \log(10)$.
$E^0_{cell} = 0.2905 + 0.02955 = 0.32005 \, V \approx 0.32 \, V$.
At equilibrium,$E_{cell} = 0$ and $E^0_{cell} = \frac{0.0591}{n} \log K_c$.
$0.32 = \frac{0.0591}{2} \log K_c$.
$0.32 = 0.02955 \log K_c$.
$\log K_c = \frac{0.32}{0.02955} \approx \frac{0.32}{0.0295}$.
$K_c = 10^{\frac{0.32}{0.0295}}$.

(C) For a concentration cell,the cell potential is given by the Nernst equation:
$E_{cell} = \frac{0.0591}{n} \log \frac{[H^+]_{cathode}}{[H^+]_{anode}}$
Here,$n = 1$,$[H^+]_{cathode} = 0.001 \ M = 10^{-3} \ M$,and $[H^+]_{anode} = 10^{-8} \ M$.
$E_{cell} = 0.0591 \log \left( \frac{10^{-3}}{10^{-8}} \right)$
$E_{cell} = 0.0591 \log(10^5)$
$E_{cell} = 0.0591 \times 5 = 0.2955 \ V$.

(C) The cell reaction is $Cu(s) + 2Ag^{+}(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s)$.
Using the Nernst equation: $E_{cell} = E^{\circ}_{cell} - \frac{0.059}{2} \log \frac{[Cu^{2+}]}{[Ag^{+}]^2}$.
When concentrations are increased by a factor of $10$,the new concentrations are $[Cu^{2+}]' = 10[Cu^{2+}]$ and $[Ag^{+}]' = 10[Ag^{+}]$.
Substituting these into the equation: $E'_{cell} = E^{\circ}_{cell} - \frac{0.059}{2} \log \frac{10[Cu^{2+}]}{(10[Ag^{+}])^2}$.
$E'_{cell} = E^{\circ}_{cell} - \frac{0.059}{2} \log \frac{10[Cu^{2+}]}{100[Ag^{+}]^2} = E^{\circ}_{cell} - \frac{0.059}{2} (\log \frac{[Cu^{2+}]}{[Ag^{+}]^2} + \log \frac{1}{10})$.
$E'_{cell} = E_{cell} - \frac{0.059}{2} (-1) = E_{cell} + 0.0295 \ V$.
Thus,the $emf$ increases by $0.0295 \ V$.

(B) The reduction half-reaction is: $Cu^{2+} + 2e^- \to Cu$
Using the Nernst equation: $E_{red} = E^o_{red} - \frac{0.059}{n} \log \frac{1}{[Cu^{2+}]}$
Here,$n = 2$ and $[Cu^{2+}] = 0.1 \, M = 10^{-1} \, M$.
Substituting the values: $E = 0.34 - \frac{0.059}{2} \log \frac{1}{10^{-1}}$
$E = 0.34 - 0.0295 \times \log(10)$
Since $\log(10) = 1$,$E = 0.34 - 0.0295 = 0.3105 \, V \approx 0.31 \, V$.

(C) The relationship between standard $emf$ and equilibrium constant $(K_{eq})$ is given by the Nernst equation at $25^o \ C$ $(298 \ K)$:
$E^o_{cell} = \frac{0.0591}{n} \log K_{eq}$
Given: $E^o_{cell} = 0.295 \ V$,$n = 2$.
Substituting the values:
$0.295 = \frac{0.0591}{2} \log K_{eq}$
$0.295 \times 2 = 0.0591 \log K_{eq}$
$0.59 = 0.0591 \log K_{eq}$
$\log K_{eq} = \frac{0.59}{0.0591} \approx 10$
Therefore,$K_{eq} = 10^{10}$.

(C) Using the Nernst equation at $298 \, K$:
$E_{cell} = E^o_{cell} - \frac{0.0591}{n} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$
Given $n = 2$,$E_{cell} = 1.3 \, V$,$[Zn^{2+}] = 0.1 \, M$,and $[Cu^{2+}] = 1.0 \, M$.
Substituting the values:
$1.3 = E^o_{cell} - \frac{0.0591}{2} \log \frac{0.1}{1.0}$
$1.3 = E^o_{cell} - 0.02955 \log(10^{-1})$
$1.3 = E^o_{cell} + 0.02955 \log(10)$
Since $\log(10) = 1$:
$1.3 = E^o_{cell} + 0.02955$
$E^o_{cell} = 1.3 - 0.02955 = 1.27045 \, V \approx 1.27 \, V$.

(A) The Nernst equation for the cell reaction is: $E_{cell} = E^o_{cell} - \frac{0.0591}{n} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$
Here,$n = 2$,$[Zn^{2+}] = 1 \ M$,and $[Cu^{2+}] = 0.1 \ M$.
Substituting the values: $E_{cell} = 1.10 - \frac{0.0591}{2} \log \frac{1}{0.1}$
$E_{cell} = 1.10 - 0.02955 \times \log(10)$
$E_{cell} = 1.10 - 0.02955 \times 1 = 1.07045 \ V \approx 1.07 \ V$.

(A) The relationship between standard $emf$ $(E^{\circ}_{cell})$ and equilibrium constant $(K_{eq})$ is given by the Nernst equation at $298 \ K$:
$E^{\circ}_{cell} = \frac{0.0591}{n} \log K_{eq}$
Given:
$E^{\circ}_{cell} = 0.295 \ V$
$n = 4$
Substituting the values:
$0.295 = \frac{0.0591}{4} \log K_{eq}$
$\log K_{eq} = \frac{0.295 \times 4}{0.0591}$
$\log K_{eq} \approx 20$
$K_{eq} = 10^{20}$

(A) The cell reaction is $Zn_{(s)} + 2H^+_{(aq)} \to Zn^{2+}_{(aq)} + H_{2(g)}$.
According to the Nernst equation: $E_{cell} = E^\circ_{cell} - \frac{0.059}{2} \log \frac{[Zn^{2+}] \times P_{H_2}}{[H^+]^2}$.
This can be rewritten as: $E_{cell} = E^\circ_{cell} + \frac{0.059}{2} \log \frac{[H^+]^2}{[Zn^{2+}] \times P_{H_2}}$.
Adding $H_2SO_4$ increases the concentration of $H^+$ ions $[H^+]$.
As $[H^+]$ increases,the logarithmic term increases,leading to an increase in $E_{cell}$.
According to Le Chatelier's principle,an increase in the concentration of reactants $(H^+)$ shifts the equilibrium to the right (towards the products).

(C) The Nernst equation is given by: $E_{cell} = E^o_{cell} - \frac{0.0591}{n} \log \left[ \frac{[Zn^{2+}]}{[Cu^{2+}]} \right]$
Here,$n = 2$,$E^o_{cell} = 1.10 \ V$,$[Zn^{2+}] = 1 \ M$,and $[Cu^{2+}] = 0.1 \ M$.
Substituting the values:
$E_{cell} = 1.10 - \frac{0.0591}{2} \log \left( \frac{1}{0.1} \right)$
$E_{cell} = 1.10 - 0.02955 \times \log(10)$
$E_{cell} = 1.10 - 0.02955 \times 1$
$E_{cell} = 1.07045 \ V \approx 1.07 \ V$.

(C) The Nernst equation is given by: $E_{cell} = E^o_{cell} - \frac{0.0592}{n} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$.
When the cell is fully discharged,$E_{cell} = 0$ and $n = 2$.
Substituting the values: $0 = 1.10 - \frac{0.0592}{2} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$.
$\frac{0.0592}{2} \log \frac{[Zn^{2+}]}{[Cu^{2+}]} = 1.10$.
$\log \frac{[Zn^{2+}]}{[Cu^{2+}]} = \frac{1.10 \times 2}{0.0592} \approx 37.16$.
Rounding to the nearest provided option,the ratio is $10^{37.3}$.

(D) The cell reaction is: $Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$.
For this reaction,$n = 2$.
The Nernst equation is: $E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log Q$,where $Q = \frac{[Zn^{2+}]}{[Cu^{2+}]}$.
At complete discharge,$E_{cell} = 0$.
Therefore,$0 = 1.10 - \frac{0.0591}{2} \log Q$.
$\log Q = \frac{1.10 \times 2}{0.0591} \approx 37.225$.
$Q = 10^{37.225} \approx 10^{37.3}$.

(C) The reduction half-reaction is: $Cu^{2+} (aq) + 2e^- \to Cu (s)$.
Using the Nernst equation: $E_{Cu^{2+}|Cu} = E^0_{Cu^{2+}|Cu} - \frac{0.0592}{n} \log \frac{1}{[Cu^{2+}]}$.
Here,$E^0 = 0.34 \, V$,$n = 2$,and $[Cu^{2+}] = 0.001 \, M = 10^{-3} \, M$.
$E_{Cu^{2+}|Cu} = 0.34 - \frac{0.0592}{2} \log \frac{1}{10^{-3}}$.
$E_{Cu^{2+}|Cu} = 0.34 - 0.0296 \times \log(10^3)$.
$E_{Cu^{2+}|Cu} = 0.34 - 0.0296 \times 3$.
$E_{Cu^{2+}|Cu} = 0.34 - 0.0888 = 0.2512 \, V$.

(D) The Nernst equation for the reduction of $Ag^+$ ions is given by:
$E_{Ag^+/Ag} = E^o_{Ag^+/Ag} - \frac{0.0591}{n} \log \frac{1}{[Ag^+]}$
For the reaction $Ag^+ + e^- \rightarrow Ag$,$n = 1$.
Substituting the values:
$E_{Ag^+/Ag} = E^o_{Ag^+/Ag} - 0.0591 \log \frac{1}{0.1}$
$E_{Ag^+/Ag} = E^o_{Ag^+/Ag} - 0.0591 \log (10)$
Since $\log (10) = 1$,we get:
$E_{Ag^+/Ag} = E^o_{Ag^+/Ag} - 0.059 \ V$
Thus,the reduction potential is $(E^o_{red} - 0.059)$.

(A) The cell reaction is: $Zn_{(s)} + Fe^{2+}_{(aq)} \rightleftharpoons Zn^{2+}_{(aq)} + Fe_{(s)}$
Using the Nernst equation: $E_{cell} = E^{\circ}_{cell} - \frac{0.059}{n} \log \frac{[Zn^{2+}]}{[Fe^{2+}]}$
Given $E_{cell} = 0.2905 \ V$,$n = 2$,$[Zn^{2+}] = 0.1 \ M$,$[Fe^{2+}] = 0.01 \ M$:
$0.2905 = E^{\circ}_{cell} - \frac{0.059}{2} \log \frac{0.1}{0.01}$
$0.2905 = E^{\circ}_{cell} - 0.0295 \log(10)$
$0.2905 = E^{\circ}_{cell} - 0.0295 \implies E^{\circ}_{cell} = 0.32 \ V$
At equilibrium,$E_{cell} = 0$,so $E^{\circ}_{cell} = \frac{0.059}{n} \log K_c$
$0.32 = \frac{0.059}{2} \log K_c$
$0.32 = 0.0295 \log K_c$
$\log K_c = \frac{0.32}{0.0295}$
$K_c = 10^{0.32/0.0295}$

(C) According to the Nernst equation,$E_{cell} = E^0_{cell} - \frac{0.0591}{n} \log \frac{[Sn^{2+}]}{[Ag^+]^2}$.
When the cell voltage $(E_{cell})$ increases,the reaction quotient $Q = \frac{[Sn^{2+}]}{[Ag^+]^2}$ must decrease.
This implies that the concentration of the product $[Sn^{2+}]$ must decrease or the concentration of the reactant $[Ag^+]$ must increase.
Therefore,an increase in the concentration of $Ag^+$ ions leads to an increase in the cell voltage.

(B) The reduction reaction is: $Cu^{+2} + 2e^- \to Cu$
According to the Nernst equation: $E_{Cu^{+2}/Cu} = E^{\circ}_{Cu^{+2}/Cu} - \frac{0.059}{2} \log \frac{1}{[Cu^{+2}]}$
When the solution is diluted $100$ times,the concentration $[Cu^{+2}]$ becomes $[Cu^{+2}]/100$. The new potential $E'$ is:
$E' = E^{\circ}_{Cu^{+2}/Cu} - \frac{0.059}{2} \log \frac{1}{[Cu^{+2}]/100}$
$E' = E^{\circ}_{Cu^{+2}/Cu} - \frac{0.059}{2} (\log \frac{1}{[Cu^{+2}]} + \log 100)$
$E' = E_{Cu^{+2}/Cu} - \frac{0.059}{2} \times 2$
$E' = E_{Cu^{+2}/Cu} - 0.059 \, V$
Thus,the electrode potential decreases by $0.059 \, V$ or $59 \, mV$.

(D) The relationship between standard cell potential $(E^{\circ}_{cell})$ and equilibrium constant $(K)$ is given by the Nernst equation at equilibrium:
$E^{\circ}_{cell} = \frac{0.0591}{n} \log K$
Given: $n = 2$,$E^{\circ}_{cell} = 0.295 \ V$.
Substituting the values:
$0.295 = \frac{0.0591}{2} \log K$
$0.295 = 0.02955 \log K$
$\log K = \frac{0.295}{0.02955} \approx 10$
$K = 10^{10} = 1 \times 10^{10}$

(A) The cell reaction is: $Zn(s) + Cu^{2+}(aq) \to Zn^{2+}(aq) + Cu(s)$.
Using the Nernst equation: $E = E^0 - \frac{0.0592}{2} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$.
For $E_1$: $[Zn^{2+}] = 0.01 \ M$ and $[Cu^{2+}] = 1 \ M$,so $E_1 = E^0 - \frac{0.0592}{2} \log(0.01) = E^0 - \frac{0.0592}{2} (-2) = E^0 + 0.0592$.
For $E_2$: $[Zn^{2+}] = 1 \ M$ and $[Cu^{2+}] = 0.01 \ M$,so $E_2 = E^0 - \frac{0.0592}{2} \log(100) = E^0 - \frac{0.0592}{2} (2) = E^0 - 0.0592$.
Comparing the two,$E_1 > E_2$.

(B) The cell reaction is: $2Cr_{(s)} + 3Fe^{2+}_{(aq)} \rightarrow 2Cr^{3+}_{(aq)} + 3Fe_{(s)}$
The standard cell potential is: $E^0_{cell} = E^0_{cathode} - E^0_{anode} = -0.42 - (-0.72) = 0.30 \, V$
Using the Nernst equation at $298 \, K$: $E_{cell} = E^0_{cell} - \frac{0.0591}{n} \log \frac{[Cr^{3+}]^2}{[Fe^{2+}]^3}$
Here,$n = 6$ (total electrons transferred).
$E_{cell} = 0.30 - \frac{0.0591}{6} \log \frac{(0.1)^2}{(0.01)^3}$
$E_{cell} = 0.30 - 0.00985 \log \frac{10^{-2}}{10^{-6}} = 0.30 - 0.00985 \log(10^4)$
$E_{cell} = 0.30 - (0.00985 \times 4) = 0.30 - 0.0394 = 0.2606 \, V \approx 0.26 \, V$

(A) The relationship between the standard cell potential $(E^{\circ}_{cell})$ and the equilibrium constant $(K_{eq})$ is given by the Nernst equation at equilibrium:
$E^{\circ}_{cell} = \frac{0.0591}{n} \log K_{eq}$
Given:
$E^{\circ}_{cell} = 0.591 \ V$
$n = 1$ (one-electron change)
Substituting the values:
$0.591 = \frac{0.0591}{1} \log K_{eq}$
$\log K_{eq} = \frac{0.591}{0.0591} = 10$
$K_{eq} = 10^{10} = 1.0 \times 10^{10}$

(B) The cell reaction is: $2Ag^{+} + Cu \to 2Ag + Cu^{2+}$
Here,the number of electrons transferred,$n = 2$.
The formula for standard Gibbs energy is: $\Delta G^{\circ} = -nFE^{\circ}_{cell}$
Substituting the values: $\Delta G^{\circ} = -2 \times 96500\,C\,mol^{-1} \times 0.46\,V$
$\Delta G^{\circ} = -88780\,J\,mol^{-1} \approx -89\,kJ\,mol^{-1}$

(A) The reduction reaction for a hydrogen half-cell is: $2H^+ (aq) + 2e^- \rightarrow H_2 (g)$.
According to the Nernst equation,the reduction potential $E_{H^+/H_2}$ is given by: $E = E^0 - \frac{0.0591}{2} \log \frac{p(H_2)}{[H^+]^2}$.
Since $E^0 = 0 \, V$,the equation becomes: $E = -0.02955 \log \frac{p(H_2)}{[H^+]^2}$.
For $E$ to be negative,$\log \frac{p(H_2)}{[H^+]^2}$ must be positive,which means $\frac{p(H_2)}{[H^+]^2} > 1$.
Checking option $A$: $\frac{2}{(1.0)^2} = 2 > 1$. Thus,$E$ is negative.

(A) The Nernst equation for the reduction half-reaction $M^{n+} + ne^- \rightarrow M$ is given by:
$E_{M^{n+}/M} = E^o_{M^{n+}/M} - \frac{RT}{nF} \ln \frac{[M]}{[M^{n+}]}$
Since the concentration of a pure solid $[M] = 1$,the equation becomes:
$E_{M^{n+}/M} = E^o_{M^{n+}/M} - \frac{0.0591}{n} \log \frac{1}{[M^{n+}]}$
This can also be written as:
$E_{M^{n+}/M} = E^o_{M^{n+}/M} + \frac{0.0591}{n} \log [M^{n+}]$

(B) The reduction reaction for a hydrogen electrode is: $2H^+ (aq) + 2e^- \rightarrow H_2 (g)$.
Using the Nernst equation: $E_{red} = E^{\circ}_{red} - \frac{0.0591}{n} \log \frac{P_{H_2}}{[H^+]^2}$.
Given $E^{\circ}_{red} = 0 \ V$,$n = 2$,$P_{H_2} = 1 \ atm$,and $[H^+] = 0.1 \ M$.
$E_{red} = 0 - \frac{0.059}{2} \log \frac{1}{(0.1)^2}$.
$E_{red} = -0.0295 \times \log(100) = -0.0295 \times 2 = -0.059 \ V$.

(A) The reduction reaction for a hydrogen electrode is: $2H^+ (aq) + 2e^- \rightarrow H_2 (g)$.
Using the Nernst equation at $298 \ K$:
$E_{H^+/H_2} = E^0_{H^+/H_2} - \frac{0.0591}{n} \log \frac{1}{[H^+]^2}$.
Given $E^0_{H^+/H_2} = 0 \ V$,$n = 2$,and $pH = -\log[H^+] = 1$,which means $[H^+] = 10^{-1} \ M$.
Substituting the values:
$E = 0 - \frac{0.0591}{2} \log \frac{1}{(10^{-1})^2} = -0.02955 \times \log(10^2) = -0.02955 \times 2 = -0.0591 \ V$.
Thus,the reduction potential is approximately $-0.059 \ V$.

(B) The cell reaction for the Daniell cell is $Zn(s) + Cu^{2+}(aq) \longrightarrow Zn^{2+}(aq) + Cu(s).$
Using the Nernst equation $E = E^{\circ} - \frac{0.059}{n} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$ where $n = 2.$
For $E_1$: $[Zn^{2+}] = 0.01 \, M$ and $[Cu^{2+}] = 1.0 \, M.$
$E_1 = E^{\circ} - \frac{0.059}{2} \log \frac{0.01}{1} = E^{\circ} - \frac{0.059}{2} \log(10^{-2}) = E^{\circ} + 0.059.$
For $E_2$: $[Zn^{2+}] = 1.0 \, M$ and $[Cu^{2+}] = 0.01 \, M.$
$E_2 = E^{\circ} - \frac{0.059}{2} \log \frac{1}{0.01} = E^{\circ} - \frac{0.059}{2} \log(10^{2}) = E^{\circ} - 0.059.$
Comparing the two,$E_1 = E^{\circ} + 0.059$ and $E_2 = E^{\circ} - 0.059.$
Therefore,$E_1 > E_2.$

(A) The relationship between standard Gibbs free energy change $(\Delta G^{\circ})$ and standard cell potential $(E^{\circ}_{cell})$ is given by $\Delta G^{\circ} = -nFE^{\circ}_{cell}$.
If $E^{\circ}_{cell} < 0$,then $\Delta G^{\circ} > 0$,which indicates that the reaction is non-spontaneous.
Furthermore,the relationship between $E^{\circ}_{cell}$ and the equilibrium constant $(K_{eq})$ is given by $E^{\circ}_{cell} = \frac{RT}{nF} \ln K_{eq}$.
If $E^{\circ}_{cell} < 0$,then $\ln K_{eq} < 0$,which implies $K_{eq} < 1$.

(C) The reduction half-reaction for the hydrogen electrode is: $2H^{+} + 2e^{-} \longrightarrow H_{2(g)}$
Using the Nernst equation for the electrode potential: $E_{H^{+}/H_2} = E^{\circ}_{H^{+}/H_2} - \frac{0.0591}{2} \log \frac{P_{H_2}}{[H^{+}]^2}$
Since $E^{\circ}_{H^{+}/H_2} = 0 \ V$ and we want the electrode potential $E_{H^{+}/H_2}$ to be $0 \ V$,we have:
$0 = 0 - \frac{0.0591}{2} \log \frac{P_{H_2}}{[H^{+}]^2}$
This implies $\log \frac{P_{H_2}}{[H^{+}]^2} = 0$,so $\frac{P_{H_2}}{[H^{+}]^2} = 10^0 = 1$
Therefore,$P_{H_2} = [H^{+}]^2$
For pure water at $298 \ K$,the concentration of hydrogen ions is $[H^{+}] = 10^{-7} \ M$
Substituting this value: $P_{H_2} = (10^{-7})^2 = 10^{-14} \ atm$

(D) The oxidation reaction for the hydrogen electrode is: $H_{2}(g) \rightarrow 2H^{+}(aq) + 2e^{-}$.
Given $pH = 10$,the concentration of hydrogen ions is $[H^{+}] = 10^{-pH} = 10^{-10} \ M$.
The pressure of hydrogen gas is $P_{H_{2}} = 1 \ atm$.
The oxidation potential is given by the Nernst equation: $E_{ox} = E_{ox}^{0} - \frac{0.0591}{n} \log Q$.
For the standard hydrogen electrode,$E_{ox}^{0} = 0 \ V$ and $n = 2$.
The reaction quotient $Q = \frac{[H^{+}]^{2}}{P_{H_{2}}} = \frac{(10^{-10})^{2}}{1} = 10^{-20}$.
Substituting the values: $E_{ox} = 0 - \frac{0.0591}{2} \log(10^{-20})$.
$E_{ox} = -\frac{0.0591}{2} \times (-20) = 0.0591 \times 10 = 0.591 \ V$.

(C) For a cell reaction in equilibrium at $298 \ K$,the relationship between the standard cell potential and the equilibrium constant is given by:
$E_{cell}^{\circ} = \frac{0.0591}{n} \log K_{C}$
where $K_{C}$ is the equilibrium constant and $n$ is the number of electrons transferred in the balanced redox reaction.
For the reaction $Cu_{(s)} + 2Ag^{+}_{(aq)} \rightarrow Cu^{2+}_{(aq)} + 2Ag_{(s)}$,the number of electrons transferred is $n = 2$.
Given $E_{cell}^{\circ} = 0.46 \ V$,we substitute the values into the equation:
$0.46 = \frac{0.0591}{2} \log K_{C}$
$\log K_{C} = \frac{0.46 \times 2}{0.0591} = \frac{0.92}{0.0591} \approx 15.567$
$K_{C} = 10^{15.567} \approx 3.69 \times 10^{15} \approx 4.0 \times 10^{15}$

(D) When the cell is completely discharged,the cell potential $E_{cell} = 0 \ V.$
The Nernst equation for the cell reaction is given by:
$E_{cell} = E_{cell}^{\circ} - \frac{0.059}{n} \log \left( \frac{[Zn^{2+}]}{[Cu^{2+}]} \right)$
Here,$n = 2$ (number of electrons transferred) and $E_{cell}^{\circ} = 1.10 \ V.$
Substituting the values:
$0 = 1.10 - \frac{0.059}{2} \log \left( \frac{[Zn^{2+}]}{[Cu^{2+}]} \right)$
Rearranging the equation:
$\log \left( \frac{[Zn^{2+}]}{[Cu^{2+}]} \right) = \frac{1.10 \times 2}{0.059} \approx 37.288 \approx 37.3$
Therefore,$\frac{[Zn^{2+}]}{[Cu^{2+}]} = 10^{37.3}$.

(D) The cell reaction is: $2Cr(s) + 3Fe^{2+}(aq) \rightarrow 2Cr^{3+}(aq) + 3Fe(s)$.
$n = 6$ (number of electrons transferred).
$E^o_{cell} = E^o_{cathode} - E^o_{anode} = -0.42 - (-0.72) = 0.30 \, V$.
Using the Nernst equation: $E_{cell} = E^o_{cell} - \frac{0.059}{n} \log \frac{[Cr^{3+}]^2}{[Fe^{2+}]^3}$.
$E_{cell} = 0.30 - \frac{0.059}{6} \log \frac{(0.1)^2}{(0.01)^3}$.
$E_{cell} = 0.30 - \frac{0.059}{6} \log \frac{10^{-2}}{10^{-6}} = 0.30 - \frac{0.059}{6} \log(10^4)$.
$E_{cell} = 0.30 - \frac{0.059}{6} \times 4 = 0.30 - 0.0393 = 0.2607 \, V \approx 0.26 \, V$.

(A) The calomel electrode reaction is: $Hg_2Cl_2(s) + 2e^{-} \rightleftharpoons 2Hg(l) + 2Cl^{-}(aq)$.
Applying the Nernst equation for the reduction potential:
$E = E^{\circ} - \frac{0.0591}{2} \log [Cl^{-}]^2$.
$E = E^{\circ} - 0.0591 \log [Cl^{-}]$.
If the concentration of $Cl^{-}$ is diluted by ten times,the new concentration is $[Cl^{-}]' = \frac{[Cl^{-}]}{10}$.
New potential $E' = E^{\circ} - 0.0591 \log \left( \frac{[Cl^{-}]}{10} \right)$.
$E' = E^{\circ} - 0.0591 (\log [Cl^{-}] - \log 10)$.
$E' = E^{\circ} - 0.0591 \log [Cl^{-}] + 0.0591$.
$E' = E + 0.0591 \ V$.
Thus,the reduction potential increases by $0.0591 \ V$ (approximately $0.059 \ V$).

(B) The Nernst equation for the given cell reaction is:
$E_{cell} = E^o_{cell} - \frac{0.059}{n} log \frac{[Zn^{2+}]}{[Cu^{2+}]}$
Here,$n = 2$,so the equation becomes:
$E_{cell} = 1.10 - \frac{0.059}{2} log \frac{[Zn^{2+}]}{[Cu^{2+}]}$
This is in the form of a linear equation $y = mx + c$,where $y = E_{cell}$,$x = log_{10} \frac{[Zn^{2+}]}{[Cu^{2+}]}$,$c = 1.10$,and the slope $m = -\frac{0.059}{2} = -0.0295$.
Since the slope is negative,the graph of $E_{cell}$ versus $log_{10} \frac{[Zn^{2+}]}{[Cu^{2+}]}$ will be a straight line with a negative slope and a $y$-intercept of $1.10 \, V$.

(B) The half-cell reaction is: $2Cr^{3+}{_{\text{(aq)}}} + 7H_2O_{\text{(l)}} \rightarrow Cr_2O_7^{2-}{_{\text{(aq)}}} + 14H^{+}{_{\text{(aq)}}} + 6e^{-}$
Using the Nernst equation: $E = E^o - \frac{0.06}{6} \log \frac{[Cr_2O_7^{2-}][H^{+}]^{14}}{[Cr^{3+}]^2}$.
$E = E^o - 0.01 \log \frac{[Cr_2O_7^{2-}]}{[Cr^{3+}]^2} - 0.01 \times 14 \log [H^{+}]$.
Since $pH = -\log[H^{+}]$,we have $E = E^o - 0.01 \log \frac{[Cr_2O_7^{2-}]}{[Cr^{3+}]^2} + 0.14 \ pH$.
The change in potential $\Delta E = E_2 - E_1 = 0.14(pH_2 - pH_1)$.
Given $pH_1 = 1$ and $pH_2 = 3$,$\Delta E = 0.14(3 - 1) = 0.14 \times 2 = 0.28 \ V$.

(A) The cell reaction is: $H_{2(g)} + 2H^{+}_{(cathode)} \rightarrow 2H^{+}_{(anode)} + H_{2(g)}$
For weak acid $HA$,$[H^{+}]_{anode} = \sqrt{K_a \times C} = \sqrt{10^{-7} \times 1} = 10^{-3.5} \ M$.
For weak acid $HB$,$[H^{+}]_{cathode} = \sqrt{K_a \times C} = \sqrt{10^{-5} \times 1} = 10^{-2.5} \ M$.
Using the Nernst equation: $E_{cell} = E^{\circ}_{cell} - \frac{0.059}{n} \log Q$.
Here,$E^{\circ}_{cell} = 0$,$n = 2$,and $Q = \frac{[H^{+}]^2_{anode}}{[H^{+}]^2_{cathode}} = \frac{(10^{-3.5})^2}{(10^{-2.5})^2} = \frac{10^{-7}}{10^{-5}} = 10^{-2}$.
$E_{cell} = 0 - \frac{0.06}{2} \log(10^{-2}) = -0.03 \times (-2) = 0.06 \ V$.

(D) The Nernst equation for the reaction is: $E = E^{\circ} - \frac{0.0591}{n} \log \frac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}][H^{+}]^{14}}$
Given $E^{\circ} = 1.33 \ V$,$n = 6$,$[Cr^{3+}] = 1 \ M$,$[Cr_2O_7^{2-}] = 1 \ M$,and $pH = 2.0$,so $[H^{+}] = 10^{-2} \ M$.
Substituting the values: $E = 1.33 - \frac{0.0591}{6} \log \frac{1^2}{1 \times (10^{-2})^{14}}$
$E = 1.33 - \frac{0.0591}{6} \log (10^{28})$
$E = 1.33 - \frac{0.0591}{6} \times 28$
$E = 1.33 - 0.2758 = 1.0542 \ V$.

(B) This is a concentration cell where the same electrode $M(s)|M^+(aq)$ is used in both half-cells.
For a concentration cell,the standard cell potential $E^0_{cell} = 0 \ V$.
The cell reaction is: $M(s) + M^+(0.1 \ M) \rightarrow M^+(0.01 \ M) + M(s)$.
Using the Nernst equation at $298 \ K$:
$E_{cell} = E^0_{cell} - \frac{0.0591}{n} \log \frac{[M^+]_{anode}}{[M^+]_{cathode}}$
Here,$n = 1$,$[M^+]_{anode} = 0.01 \ M$,and $[M^+]_{cathode} = 0.1 \ M$.
$E_{cell} = 0 - \frac{0.0591}{1} \log \frac{0.01}{0.1}$
$E_{cell} = -0.0591 \times \log(0.1)$
$E_{cell} = -0.0591 \times (-1) = 0.0591 \ V \approx 0.059 \ V$.

(B) The cell reaction is: $M + nH^{+} \longrightarrow M^{n+} + \frac{n}{2} H_{2}$.
The standard cell potential $E^{\circ}_{cell} = E^{\circ}_{ox}(M) + E^{\circ}_{red}(H^{+}/H_{2}) = 0.76 \, V + 0 \, V = 0.76 \, V$.
Using the Nernst equation: $E_{cell} = E^{\circ}_{cell} - \frac{0.06}{n} \log \frac{[M^{n+}]}{[H^{+}]^{n}}$.
Given $E_{cell} = 0.81 \, V$,$[M^{n+}] = 0.02 \, M$,and $[H^{+}] = 1 \, M$:
$0.81 = 0.76 - \frac{0.06}{n} \log \frac{0.02}{(1)^{n}}$.
$0.05 = -\frac{0.06}{n} \log (2 \times 10^{-2})$.
$0.05 = -\frac{0.06}{n} (\log 2 + \log 10^{-2}) = -\frac{0.06}{n} (0.3 - 2) = -\frac{0.06}{n} (-1.7)$.
$0.05 = \frac{0.102}{n} \implies n = \frac{0.102}{0.05} \approx 2$.

(A) The cell reaction is $Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$.
$E^0_{cell} = E^0_{cathode} - E^0_{anode} = 0.34 - (-0.76) = 1.10\ V$.
At $emf = 1.10\ V$,$E_{cell} = E^0_{cell} - \frac{0.06}{2} \log \frac{[Zn^{2+}]}{[Cu^{2+}]} = 1.10\ V$.
This implies $\log \frac{[Zn^{2+}]}{[Cu^{2+}]} = 0$,so $[Zn^{2+}] = [Cu^{2+}]$.
Let $x$ be the amount of $Zn$ oxidized. $[Zn^{2+}] = 0.1 + x$ and $[Cu^{2+}] = 0.9 - x$.
$0.1 + x = 0.9 - x$ $\Rightarrow 2x = 0.8$ $\Rightarrow x = 0.4$.
So,$[Zn^{2+}] = 0.5\ M$ and $[Cu^{2+}] = 0.5\ M$. Option $A$ is incorrect as $[Zn^{2+}]$ becomes $0.5\ M$,not $1\ M$.
After charging with $0.6\ F$,$0.3\ mol$ of $Zn^{2+}$ is reduced to $Zn$ and $0.3\ mol$ of $Cu$ is oxidized to $Cu^{2+}$.
New $[Zn^{2+}] = 0.5 - 0.3 = 0.2\ M$ and $[Cu^{2+}] = 0.5 + 0.3 = 0.8\ M$.
$E_{cell} = 1.10 - 0.03 \log \frac{0.2}{0.8} = 1.10 - 0.03 \log(0.25) = 1.10 - 0.03(-0.6) = 1.10 + 0.018 = 1.118\ V$.

(A) The balanced cell reaction is: $10Fe^{+2} + 2MnO_4^- + 16H^{+} \to 2Mn^{+2} + 10Fe^{+3} + 8H_2O$
The number of electrons transferred $(n)$ is $10$.
The reaction quotient $Q$ is given by: $Q = \frac{[Mn^{+2}]^2 [Fe^{+3}]^{10}}{[Fe^{+2}]^{10} [MnO_4^-]^2 [H^{+}]^{16}}$
Given $[H^+] = 10^{-pH} = 10^{-1} \ M$. Substituting the values:
$Q = \frac{(10^{-2})^2 (10^{-3})^{10}}{(10^{-2})^{10} (10^{-3})^2 (10^{-1})^{16}} = \frac{10^{-4} \times 10^{-30}}{10^{-20} \times 10^{-6} \times 10^{-16}} = \frac{10^{-34}}{10^{-42}} = 10^8$
Using the Nernst equation: $E_{cell} = E^o_{cell} - \frac{0.0591}{n} \log(Q)$
$E_{cell} = -2.31 - \frac{0.0591}{10} \log(10^8)$
$E_{cell} = -2.31 - 0.00591 \times 8 = -2.31 - 0.04728 = -2.35728 \ V \approx -2.36 \ V$

(B) The cell reaction is: $AgBr_{(s)} + I^-_{(aq)} \rightleftharpoons AgI_{(s)} + Br^-_{(aq)}$
The equilibrium constant $K_{eq}$ is given by: $K_{eq} = \frac{K_{sp}(AgBr)}{K_{sp}(AgI)} = \frac{4 \times 10^{-13}}{8 \times 10^{-17}} = 0.5 \times 10^4 = 5000$.
$E^o_{cell} = \frac{0.06}{1} \log K_{eq} = 0.06 \times \log(5000) = 0.06 \times (3 + \log 5) = 0.06 \times (3 + 0.7) = 0.06 \times 3.7 = 0.222 \ V$.
Using the Nernst equation: $E_{cell} = E^o_{cell} - 0.06 \log \frac{[Br^-]}{[I^-]} = 0.222 - 0.06 \log \frac{0.01}{0.02} = 0.222 - 0.06 \log(0.5) = 0.222 - 0.06(-0.3) = 0.222 + 0.018 = 0.240 \ V$.
$\Delta G^o = -nFE^o_{cell} = -1 \times 96500 \times 0.222 \ J$.

(C) For simultaneous deposition,the reduction potentials of both ions must be equal at the cathode.
Using the Nernst equation for both half-reactions:
$E_{Cu^{2+}/Cu} = E^0_{Cu^{2+}/Cu} + \frac{0.0591}{2} \log [Cu^{2+}]$
$E_{Ag^{+}/Ag} = E^0_{Ag^{+}/Ag} + 0.0591 \log [Ag^{+}]$
Equating the two potentials:
$0.34 + \frac{0.0591}{2} \log(1.0) = 0.812 + 0.0591 \log [Ag^{+}]$
Since $\log(1.0) = 0$,the equation becomes:
$0.34 = 0.812 + 0.0591 \log [Ag^{+}]$
$0.0591 \log [Ag^{+}] = 0.34 - 0.812 = -0.472$
$\log [Ag^{+}] = \frac{-0.472}{0.0591} \approx -7.986 \approx -8$
$[Ag^{+}] = 10^{-8} \ M$

(D) The cell reaction for a Daniell cell is $Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$.
If the reaction is multiplied by $2$,it becomes $2Zn(s) + 2Cu^{2+}(aq) \rightarrow 2Zn^{2+}(aq) + 2Cu(s)$.
The Nernst equation is $E_{cell} = E_{cell}^0 - \frac{RT}{nF} \ln Q$.
For the original reaction,$n=2$ and $Q = \frac{[Zn^{2+}]}{[Cu^{2+}]}$.
For the multiplied reaction,$n'=4$ and $Q' = \frac{[Zn^{2+}]^2}{[Cu^{2+}]^2} = Q^2$.
Substituting into the Nernst equation: $E'_{cell} = E_{cell}^0 - \frac{RT}{4F} \ln(Q^2) = E_{cell}^0 - \frac{RT}{4F} (2 \ln Q) = E_{cell}^0 - \frac{RT}{2F} \ln Q = E_{cell}$.
Thus,$E_{cell}$ and $E_{cell}^0$ remain unchanged.
However,$\ln Q' = \ln(Q^2) = 2 \ln Q$. Therefore,the value of $\ln Q$ doubles.

(B) The Nernst equation for the half-cell $Cu^{2+} + 2e^- \rightarrow Cu$ is:
$E = E^0 - \frac{0.0591}{2} \log \frac{1}{[Cu^{2+}]}$
After dilution to $100$ times,the concentration $[Cu^{2+}]$ becomes $\frac{[Cu^{2+}]}{100}$.
New potential $E' = E^0 - \frac{0.0591}{2} \log \frac{1}{[Cu^{2+}]/100} = E^0 - \frac{0.0591}{2} (\log \frac{1}{[Cu^{2+}]} + \log 100)$
$E' = E - \frac{0.0591}{2} \times 2 = E - 0.0591 \ V = E - 59 \ mV$.
Therefore,the potential decreases by $59 \ mV$.

(C) The cell reaction is: $H_2(P_1) + 2H^{+}(X_2) \rightarrow 2H^{+}(X_1) + H_2(P_2)$.
Using the Nernst equation for the cell potential $E_{cell}$:
$E_{cell} = E^0_{cell} - \frac{0.0591}{2} \log \frac{[H^{+}]^2_{X_1} \cdot P_2}{[H^{+}]^2_{X_2} \cdot P_1}$.
Since $E^0_{cell} = 0$ for a concentration cell,$E_{cell} = -0.02955 \log \frac{X_1^2 \cdot P_2}{X_2^2 \cdot P_1}$.
For the reaction to be spontaneous,$E_{cell} > 0$,which implies $\log \frac{X_1^2 \cdot P_2}{X_2^2 \cdot P_1} < 0$.
This means $\frac{X_1^2 \cdot P_2}{X_2^2 \cdot P_1} < 1$,or $\frac{X_1^2}{X_2^2} < \frac{P_1}{P_2}$.
If $X_1 = X_2$,then $1 < \frac{P_1}{P_2}$,which implies $P_2 < P_1$ or $P_1 > P_2$.

(A) The oxidation half-cell reaction is: $Hydroquinone \rightleftharpoons Quinone + 2H^+ + 2e^-$
Using the Nernst equation for the oxidation potential $(E_{op})$:
$E_{op} = E_{op}^o - \frac{0.059}{n} \log_{10} Q$
Here,$n = 2$ and $Q = [H^+]^2$ (assuming the ratio of quinone to hydroquinone is $1$ in quinhydrone electrode).
$E_{op} = E_{op}^o - \frac{0.059}{2} \log_{10} [H^+]^2$
$E_{op} = E_{op}^o - 0.059 \log_{10} [H^+]$
Since $pH = -\log_{10} [H^+]$,we have:
$E_{op} = E_{op}^o + 0.059 \times pH$
Given $E_{op}^o = 1.30 \ V$ and $pH = 3$:
$E_{op} = 1.30 + 0.059 \times 3$
$E_{op} = 1.30 + 0.177$
$E_{op} = 1.477 \ V \approx 1.48 \ V$