Nernst equation and ECS Questions in English

(C) The cell reaction is: $Mg_{(s)} + Cu^{2+}_{(aq)} \rightarrow Mg^{2+}_{(aq)} + Cu_{(s)}$
The Nernst equation is: $E_{cell} = E^0_{cell} - \frac{RT}{nF} \ln(Q)$
Here,$n = 2$,$E^0_{cell} = 2.70 \ V$,$T = 300 \ K$,and $Q = \frac{[Mg^{2+}]}{[Cu^{2+}]}$.
Given $\frac{F}{R} = 11500 \ K \ V^{-1}$,so $\frac{R}{F} = \frac{1}{11500} \ V \ K^{-1}$.
Substituting the values into the Nernst equation:
$2.67 = 2.70 - \frac{300}{2 \times 11500} \ln \left( \frac{x}{1} \right)$
$-0.03 = -\frac{300}{23000} \ln(x)$
$0.03 = \frac{3}{230} \ln(x)$
$\ln(x) = 0.03 \times \frac{230}{3} = 0.01 \times 230 = 2.30$
Since $\ln(10) = 2.30$,we get $x = 10$.

(D) The cell reaction is: $X_{(s)} + Y^{2+}(0.1 \ M) \longrightarrow X^{2+}(0.001 \ M) + Y_{(s)}$
According to the Nernst equation: $E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.0591}{2} \log \frac{[X^{2+}]}{[Y^{2+}]}$
Here,$\frac{[X^{2+}]}{[Y^{2+}]} = \frac{0.001}{0.1} = 0.01 = 10^{-2}$
Therefore,$E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.0591}{2} \log(10^{-2}) = E^{\circ}_{\text{cell}} + 0.0591 \approx E^{\circ}_{\text{cell}} + 0.06 \ V$
Since $X$ undergoes dissolution,$X$ acts as the anode. For the reaction to be spontaneous,$E_{\text{cell}} > 0$.
$E^{\circ}_{\text{cell}} = E^{\circ}_{Y^{2+}/Y} - E^{\circ}_{X^{2+}/X}$
$(A) \ Cd$ and $Ni: E^{\circ}_{\text{cell}} = -0.24 - (-0.40) = +0.16 \ V; E_{\text{cell}} = +0.16 + 0.06 = +0.22 \ V > 0$
$(B) \ Cd$ and $Fe: E^{\circ}_{\text{cell}} = -0.44 - (-0.40) = -0.04 \ V; E_{\text{cell}} = -0.04 + 0.06 = +0.02 \ V > 0$
$(C) \ Ni$ and $Pb: E^{\circ}_{\text{cell}} = -0.13 - (-0.24) = +0.11 \ V; E_{\text{cell}} = +0.11 + 0.06 = +0.17 \ V > 0$
$(D) \ Ni$ and $Fe: E^{\circ}_{\text{cell}} = -0.44 - (-0.24) = -0.20 \ V; E_{\text{cell}} = -0.20 + 0.06 = -0.14 \ V < 0$
The reaction is spontaneous when $E_{\text{cell}} > 0$. Thus,combinations $(A), (B),$ and $(C)$ are correct.

(A) The cell reaction for $Mg|Mg^{2+}_{(aq)}||Ag^{+}_{(aq)}|Ag$ is $:$
$Mg_{(s)} + 2Ag^{+}_{(aq)} \rightarrow Mg^{2+}_{(aq)} + 2Ag_{(s)}$
Here,the number of electrons transferred $n = 2$.
The reaction quotient $Q$ is given by $Q = \frac{[Mg^{2+}]}{[Ag^{+}]^2}$.
According to the Nernst equation $:$
$E_{\text{cell}} = E_{\text{cell}}^{o} - \frac{RT}{nF} \ln Q$
Substituting the values $:$
$E_{\text{cell}} = E_{\text{cell}}^{o} - \frac{RT}{2F} \ln \frac{[Mg^{2+}]}{[Ag^{+}]^2}$
Thus,the correct option is $A$.

(A) $1$. For $Cu^{2+} + 2e^{-} \longrightarrow Cu$:
Initial moles of $Cu^{2+} = 1.5 \ M \times 1 \ L = 1.5 \ mol$.
$1 \ F$ electricity corresponds to $1 \ mol$ of electrons.
Since $2 \ mol$ of electrons reduce $1 \ mol$ of $Cu^{2+}$,$1 \ mol$ of electrons reduces $0.5 \ mol$ of $Cu^{2+}$.
Remaining moles of $Cu^{2+} = 1.5 - 0.5 = 1.0 \ mol$.
$[Cu^{2+}] = 1.0 \ M$.
$2$. For $Ag^{+} + e^{-} \longrightarrow Ag$:
Initial moles of $Ag^{+} = 0.2 \ M \times 1 \ L = 0.2 \ mol$.
$0.1 \ F$ electricity corresponds to $0.1 \ mol$ of electrons.
$1 \ mol$ of electrons reduces $1 \ mol$ of $Ag^{+}$,so $0.1 \ mol$ of electrons reduces $0.1 \ mol$ of $Ag^{+}$.
Remaining moles of $Ag^{+} = 0.2 - 0.1 = 0.1 \ mol$.
$[Ag^{+}] = 0.1 \ M$.
$3$. Cell reaction: $Cu(s) + 2Ag^{+}(aq) \longrightarrow Cu^{2+}(aq) + 2Ag(s)$.
$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 0.8 - 0.34 = 0.46 \ V$.
Using Nernst equation: $E = E^{\circ}_{cell} - \frac{0.06}{n} \log \frac{[Cu^{2+}]}{[Ag^{+}]^2}$.
$E = 0.46 - \frac{0.06}{2} \log \frac{1}{(0.1)^2} = 0.46 - 0.03 \log(100) = 0.46 - 0.03 \times 2 = 0.46 - 0.06 = 0.40 \ V$.

(A) The Nernst equation for the given half-cell reaction is: $E = E^{o} - \frac{0.059}{n} \log Q$.
Here,$n = 6$ and $Q = \frac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}] [H^{+}]^{14}}$.
Setting $E = 0$:
$0 = 1.33 - \frac{0.059}{6} \log \left( \frac{10^{-6}}{[H^{+}]^{14}} \right)$.
$1.33 = \frac{0.059}{6} (\log 10^{-6} - \log [H^{+}]^{14})$.
$1.33 = \frac{0.059}{6} (-6 + 14 pH)$.
$\frac{1.33 \times 6}{0.059} = -6 + 14 pH$.
$135.25 = -6 + 14 pH$.
$141.25 = 14 pH$.
$pH = \frac{141.25}{14} \approx 10.09$.
The nearest integer value is $10$.

(B) The cell reaction is: $H_2(P_1) \rightarrow 2H^{+} + 2e^{-}$ (Anode) and $2H^{+} + 2e^{-} \rightarrow H_2(P_2)$ (Cathode).
Overall reaction: $H_2(P_1) \rightarrow H_2(P_2)$.
Using the Nernst equation: $E_{cell} = E^{\circ}_{cell} - \frac{RT}{nF} \ln Q$.
Here,$E^{\circ}_{cell} = 0 \ V$,$n = 2$,and $Q = \frac{P_2}{P_1}$.
$E_{cell} = 0 - \frac{RT}{2F} \ln \frac{P_2}{P_1}$.
$E_{cell} = \frac{RT}{2F} \ln \frac{P_1}{P_2}$.

(A) The standard cell potential $E^0_{cell}$ is calculated as $E^0_{cathode} - E^0_{anode}$.
$E^0_{cell} = E^0_{Fe^{2+}/Fe} - E^0_{Cr^{3+}/Cr} = -0.45 \ V - (-0.75 \ V) = 0.30 \ V$.
The Nernst equation at $298 \ K$ is $E_{cell} = E^0_{cell} - \frac{0.0591}{n} \log \frac{[Cr^{3+}]^2}{[Fe^{2+}]^3}$.
Here,$n = 6$ (number of electrons transferred).
$E_{cell} = 0.30 - \frac{0.0591}{6} \log \frac{(0.1)^2}{(0.01)^3} = 0.30 - 0.00985 \log \frac{10^{-2}}{10^{-6}} = 0.30 - 0.00985 \log(10^4)$.
$E_{cell} = 0.30 - 0.00985 \times 4 = 0.30 - 0.0394 = 0.2606 \ V$.

(A) The relationship between the standard cell potential $(E_{cell}^{\circ})$ and the equilibrium constant $(K_c)$ is given by the formula: $E_{cell}^{\circ} = \frac{0.0591}{n} \log K_c$.
Given $\frac{2.303 \ RT}{F} = 0.06$,the formula becomes $E_{cell}^{\circ} = \frac{0.06}{n} \log K_c$.
In the reaction $Ni_{(s)} + 2Ag_{(aq)}^{+} \rightarrow Ni_{(aq)}^{2+} + 2Ag_{(s)}$,the number of electrons transferred $(n)$ is $2$.
Substituting the values: $1.05 = \frac{0.06}{2} \log K_c$.
$1.05 = 0.03 \log K_c$.
$\log K_c = \frac{1.05}{0.03} = 35$.
$K_c = 10^{35}$.

(C) The relationship between the standard $e.m.f.$ $(E^{\circ}_{cell})$ and the equilibrium constant $(K_{eq})$ is given by the Nernst equation at equilibrium:
$E^{\circ}_{cell} = \frac{2.303 RT}{nF} \log K_{eq}$
Given:
$E^{\circ}_{cell} = 0.591 \ V$
$n = 1$
$T = 25^{\circ} C = 298 \ K$
At $298 \ K$,the value of $\frac{2.303 RT}{F} \approx 0.0591 \ V$.
Substituting the values:
$0.591 = \frac{0.0591}{1} \log K_{eq}$
$\log K_{eq} = \frac{0.591}{0.0591} = 10$
$K_{eq} = 10^{10} = 1.0 \times 10^{10}$
Therefore,the correct option is $C$.

(C) The Nernst equation for the given cell reaction is: $E_{cell} = E_{cell}^0 - \frac{0.059}{2} \log \frac{[Sn^{2+}]}{[Ag^{+}]^2}$.
According to the equation,$E_{cell}$ depends on the ratio of the concentrations of the products to the reactants.
To increase the voltage $(E_{cell})$,we need to decrease the value of the logarithmic term $\frac{[Sn^{2+}]}{[Ag^{+}]^2}$.
This can be achieved by increasing the concentration of the reactant $Ag^{+}$ ions or by decreasing the concentration of the product $Sn^{2+}$ ions.
Therefore,an increase in the concentration of $Ag^{+}$ ions will increase the voltage of the cell.

(C) The cell reaction is: $2Cr + 3Fe^{2+} \rightarrow 2Cr^{3+} + 3Fe$. The number of electrons transferred is $n = 6$.
$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = -0.45 \ V - (-0.75 \ V) = 0.30 \ V$.
Using the Nernst equation: $E_{cell} = E^{\circ}_{cell} - \frac{0.059}{n} \log \frac{[Cr^{3+}]^2}{[Fe^{2+}]^3}$.
$E_{cell} = 0.30 - \frac{0.059}{6} \log \frac{(0.1)^2}{(0.01)^3}$.
$E_{cell} = 0.30 - \frac{0.059}{6} \log \frac{10^{-2}}{10^{-6}} = 0.30 - \frac{0.059}{6} \log(10^4)$.
$E_{cell} = 0.30 - \frac{0.059 \times 4}{6} = 0.30 - 0.0393 \approx 0.26 \ V$.

(A) The Nernst equation for the cell reaction is given by: $E_{cell} = E^{\circ}_{cell} - \frac{0.0592}{n} \log Q$,where $n = 2$ and $Q = \frac{[Zn^{2+}]}{[Ag^{+}]^2}$.
Given that $E_{cell} = E^{\circ}_{cell} - 0.0592 \ V$,we have:
$E^{\circ}_{cell} - \frac{0.0592}{2} \log Q = E^{\circ}_{cell} - 0.0592$
$\frac{1}{2} \log Q = 1 \implies \log Q = 2 \implies Q = 10^2 = 100$.
Checking the options:
For option $A$: $Q = \frac{1}{(0.1)^2} = \frac{1}{0.01} = 100$.
Thus,option $A$ is correct.

(C) The cell reaction is: $Zn_{(s)} + 2Ag^{+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + 2Ag_{(s)}$.
Using the Nernst equation: $E_{cell} = E^{\circ}_{cell} - \frac{0.0592}{n} \log Q$.
Here,$n = 2$ and $Q = \frac{[Zn^{2+}]}{[Ag^{+}]^2}$.
Initially,$Q_1 = \frac{1}{(1)^2} = 1$,so $E_1 = E^{\circ}_{cell} - \frac{0.0592}{2} \log(1) = E^{\circ}_{cell}$.
After the concentration change,$Q_2 = \frac{0.1}{(1)^2} = 0.1$.
$E_2 = E^{\circ}_{cell} - \frac{0.0592}{2} \log(0.1) = E^{\circ}_{cell} - 0.0296 \times (-1) = E^{\circ}_{cell} + 0.0296 \ V$.
The change in $EMF$ is $E_2 - E_1 = +0.0296 \ V$.
Thus,the $EMF$ increases by $0.0296 \ V$.

(D) The Nernst equation for the given cell reaction is: $E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log \frac{[Cd^{2+}]}{[Cu^{2+}]}$.
Here,the number of electrons transferred $n = 2$.
Given that $[Cd^{2+}] = 10 [Cu^{2+}]$,so the ratio $\frac{[Cd^{2+}]}{[Cu^{2+}]} = 10$.
Substituting these values into the equation: $E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{2} \log(10)$.
Since $\log(10) = 1$,we get $E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{2} \times 1$.
$E_{cell} = E^{\circ}_{cell} - 0.02955 \ V \approx E^{\circ}_{cell} - 0.0296 \ V$.
Therefore,the $emf$ of the cell is less than $E^{\circ}_{cell}$ by $0.0296 \ V$.

(A) The given reaction is the oxidation of $Mg$ to $Mg^{2+}$.
The standard electrode potential for reduction is $E^{\circ} (Mg^{2+} \mid Mg) = -2.37 \ V$.
Therefore,the standard oxidation potential is $E^{\circ}_{ox} = -(-2.37 \ V) = +2.37 \ V$.
Using the Nernst equation for the reaction $Mg_{(s)} \longrightarrow Mg^{2+} (0.1 \ M) + 2 \ e^{-}$:
$E = E^{\circ}_{ox} - \frac{0.0591}{n} \log [Mg^{2+}]$
$E = 2.37 - \frac{0.0591}{2} \log (0.1)$
$E = 2.37 - 0.02955 \times (-1)$
$E = 2.37 + 0.02955 = 2.39955 \ V \approx +2.3996 \ V$.

(D) The given half-reaction is the oxidation of silver: $Ag_{(s)} \rightarrow Ag^{+}_{(aq)} + e^{-}$.
Given $E^{\circ}(Ag^{+}/Ag) = +0.80 \ V$,the standard oxidation potential is $E^{\circ}_{ox} = -E^{\circ}_{red} = -0.80 \ V$.
Using the Nernst equation for the oxidation half-cell at $298 \ K$:
$E = E^{\circ}_{ox} - \frac{0.0591}{n} \log [Ag^{+}]$
Here,$n = 1$ and $[Ag^{+}] = 0.01 \ M = 10^{-2} \ M$.
$E = -0.80 - \frac{0.0591}{1} \log(10^{-2})$
$E = -0.80 - 0.0591 \times (-2)$
$E = -0.80 + 0.1182$
$E = -0.6818 \ V \approx -0.6816 \ V$.

(C) The given reaction is the oxidation of $Mg$ to $Mg^{+2}$. The standard reduction potential is $E^{\circ}(Mg^{+2} \mid Mg) = -2.37 \ V$. Therefore,the standard oxidation potential is $E^{\circ}_{ox} = -(-2.37 \ V) = +2.37 \ V$.
Using the Nernst equation for the oxidation half-cell: $E_{ox} = E^{\circ}_{ox} - \frac{0.0591}{n} \log [Mg^{+2}]$.
Here,$n = 2$ and $[Mg^{+2}] = 0.01 \ M = 10^{-2} \ M$.
Substituting the values: $E_{ox} = 2.37 - \frac{0.0591}{2} \log(10^{-2})$.
$E_{ox} = 2.37 - 0.02955 \times (-2)$.
$E_{ox} = 2.37 + 0.0591 = 2.4291 \ V$.
Rounding to four decimal places,we get $2.4292 \ V$.

(C) The given half-cell reaction is the oxidation of aluminum: $Al_{(s)} \rightarrow Al_{(aq)}^{+3} + 3e^-$.
The standard reduction potential is $E^{\circ}(Al^{+3} \mid Al) = -1.66 \ V$,so the standard oxidation potential is $E^{\circ}_{ox} = +1.66 \ V$.
Using the Nernst equation for the oxidation potential: $E_{ox} = E^{\circ}_{ox} - \frac{0.0591}{n} \log[Al^{+3}]$.
Here,$n = 3$ and $[Al^{+3}] = 0.1 \ M$.
$E_{ox} = 1.66 - \frac{0.0591}{3} \log(0.1)$.
$E_{ox} = 1.66 - 0.0197 \times (-1)$.
$E_{ox} = 1.66 + 0.0197 = 1.6797 \ V \approx +1.679 \ V$.

(A) The Nernst equation at $298 \ K$ is given by: $E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log \frac{[Zn^{+2}]}{[Cu^{+2}]}$.
Here,$n = 2$ (number of electrons transferred).
Given $[Zn^{+2}] = 0.1 \ M$ and $[Cu^{+2}] = 0.1 \ M$.
Substituting the values: $E_{cell} = 1.1 - \frac{0.0591}{2} \log \frac{0.1}{0.1}$.
Since $\log(1) = 0$,the equation becomes: $E_{cell} = 1.1 - 0 = 1.1 \ V$.

(D) The cell reaction is: $Cd_{(s)} + Cu^{2+}_{(aq)} \rightarrow Cd^{2+}_{(aq)} + Cu_{(s)}$.
Using the Nernst equation: $E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log \frac{[Cd^{2+}]}{[Cu^{2+}]}$.
Here,$n = 2$,$E^{\circ}_{cell} = 0.74 \ V$.
Given that both concentrations decrease by $10$ times,the ratio $\frac{[Cd^{2+}]}{[Cu^{2+}]}$ remains $\frac{0.1}{0.1} = 1$.
Therefore,$E_{cell} = 0.74 - \frac{0.0591}{2} \log(1)$.
Since $\log(1) = 0$,$E_{cell} = 0.74 \ V$.

(A) The cell reaction is: $Ni_{(s)} + 2Ag^{+}_{(aq)} \rightarrow Ni^{2+}_{(aq)} + 2Ag_{(s)}$.
Applying the Nernst equation at $298 \ K$: $E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log Q$.
Here,$n = 2$ and $Q = \frac{[Ni^{2+}]}{[Ag^{+}]^2} = \frac{0.01}{(0.01)^2} = \frac{0.01}{0.0001} = 100$.
Substituting the values: $E_{cell} = E^{\circ}_{cell} - \frac{0.0592}{2} \log(100)$.
$E_{cell} = E^{\circ}_{cell} - 0.0296 \times 2 = E^{\circ}_{cell} - 0.0592 \ V$.
Thus,the cell emf is less than $E^{\circ}_{cell}$ by $0.0592 \ V$.

(A) The given reaction is the oxidation half-reaction: $Zn_{(s)} \rightarrow Zn_{(0.01 \ M)}^{+2} + 2e^-$.
The standard reduction potential is $E^{\circ}(Zn^{+2} \mid Zn) = -0.76 \ V$.
Therefore,the standard oxidation potential is $E^{\circ}_{ox} = -E^{\circ}_{red} = -(-0.76 \ V) = +0.76 \ V$.
Using the Nernst equation for the oxidation half-cell:
$E_{ox} = E^{\circ}_{ox} - \frac{0.0591}{n} \log([Zn^{+2}])$.
Here,$n = 2$ and $[Zn^{+2}] = 0.01 \ M = 10^{-2} \ M$.
$E_{ox} = 0.76 - \frac{0.0591}{2} \log(10^{-2})$.
$E_{ox} = 0.76 - 0.02955 \times (-2)$.
$E_{ox} = 0.76 + 0.0591 = 0.8191 \ V \approx +0.8192 \ V$.

(D) The given reaction is the oxidation of copper: $Cu_{(s)} \rightarrow Cu^{2+}_{(aq)} (0.1 \ M) + 2e^-$.
The standard reduction potential is $E^{\circ}(Cu^{2+} \mid Cu) = +0.34 \ V$.
The standard oxidation potential is $E^{\circ}_{ox} = -E^{\circ}_{red} = -0.34 \ V$.
Using the Nernst equation for the oxidation half-cell:
$E_{ox} = E^{\circ}_{ox} - \frac{0.0591}{n} \log [Cu^{2+}]$.
Here,$n = 2$ and $[Cu^{2+}] = 0.1 \ M$.
$E_{ox} = -0.34 - \frac{0.0591}{2} \log(0.1)$.
$E_{ox} = -0.34 - 0.02955 \times (-1)$.
$E_{ox} = -0.34 + 0.02955 = -0.31045 \ V \approx -0.3104 \ V$.

(B) The relationship between the standard $EMF$ of the cell $(E^{\circ}_{cell})$ and the equilibrium constant $(K_c)$ is given by the Nernst equation at equilibrium:
$E^{\circ}_{cell} = \frac{0.0591}{n} \log K_c$
Here,the number of electrons transferred $(n)$ is $2$ as $A \rightarrow A^{2+} + 2e^-$ and $B^{2+} + 2e^- \rightarrow B$.
Given $K_c = 10^4$ and $n = 2$.
Substituting the values:
$E^{\circ}_{cell} = \frac{0.0591}{2} \log(10^4)$
$E^{\circ}_{cell} = \frac{0.0591}{2} \times 4$
$E^{\circ}_{cell} = 0.0591 \times 2 = 0.1182 \ V$.
Rounding to the nearest option,we get $0.1184 \ V$.

(B) The cell reaction is: $Zn_{(s)} + 2Ag^{+1}_{(aq)} \rightarrow Zn^{+2}_{(aq)} + 2Ag_{(s)}$.
Using the Nernst equation: $E_{cell} = E^0_{cell} - \frac{0.0592}{n} \log Q$.
Here,$n = 2$ and $Q = \frac{[Zn^{+2}]}{[Ag^{+1}]^2}$.
Initially,$[Zn^{+2}] = 1 \ M$ and $[Ag^{+1}] = 1 \ M$,so $Q_1 = \frac{1}{1^2} = 1$. Thus,$E_1 = E^0_{cell} - \frac{0.0592}{2} \log(1) = E^0_{cell}$.
Finally,$[Zn^{+2}] = 1 \ M$ and $[Ag^{+1}] = 0.1 \ M$,so $Q_2 = \frac{1}{(0.1)^2} = \frac{1}{0.01} = 100$.
$E_2 = E^0_{cell} - \frac{0.0592}{2} \log(100) = E^0_{cell} - 0.0296 \times 2 = E^0_{cell} - 0.0592 \ V$.
The change in $emf$ is $E_2 - E_1 = -0.0592 \ V$,which means the $emf$ decreases by $0.0592 \ V$.

(B) The reduction half-reaction for the hydrogen electrode is: $2H^+ (aq) + 2e^- \rightarrow H_2 (g)$.
According to the Nernst equation at $298 \ K$:
$E_{red} = E^\circ_{red} - \frac{0.0591}{n} \log \frac{P_{H_2}}{[H^+]^2}$.
Given $E^\circ_{red} = 0 \ V$,$P_{H_2} = 1 \ atm$,$n = 2$,and $pH = 1$ (which means $[H^+] = 10^{-1} \ M$):
$E_{red} = 0 - \frac{0.0591}{2} \log \frac{1}{(10^{-1})^2}$.
$E_{red} = -0.02955 \times \log(10^2) = -0.02955 \times 2 = -0.0591 \ V$.
Rounding to the nearest option,the value is $-0.0592 \ V$.

(A) The cell reaction is: $Pb_{(s)} + 2Ag^{+}_{(10 \ M)} \rightarrow Pb^{2+}_{(1 \ M)} + 2Ag_{(s)}$
Using the Nernst equation at $25^{\circ} C$ $(298 \ K)$:
$E_{cell} = E^{\circ}_{cell} - \frac{0.0592}{n} \log_{10} Q$
Here,$n = 2$ (number of electrons transferred).
$Q = \frac{[Pb^{2+}]}{[Ag^{+}]^{2}} = \frac{1}{(10)^{2}} = 10^{-2}$
Substituting the values:
$E_{cell} = E^{\circ}_{cell} - \frac{0.0592}{2} \log_{10} (10^{-2})$
$E_{cell} = E^{\circ}_{cell} - \frac{0.0592}{2} \times (-2)$
$E_{cell} = E^{\circ}_{cell} + 0.0592 \ V$

(D) The cell reaction is: $Cd_{(s)} + 2Ag^{+}_{(aq)} \longrightarrow Cd^{2+}_{(aq)} + 2Ag_{(s)}$
For this reaction,the number of electrons transferred is $n = 2$.
The Nernst equation at $298 \ K$ is given by: $E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.0592}{n} \log_{10} \frac{[Cd^{2+}]}{[Ag^{+}]^2}$
Given $[Cd^{2+}] = 1 \ M$ and $[Ag^{+}] = 1 \ M$,the reaction quotient $Q = \frac{1}{(1)^2} = 1$.
Substituting the values: $E_{\text{cell}} = 1.2 - \frac{0.0592}{2} \log_{10}(1)$
Since $\log_{10}(1) = 0$,we get $E_{\text{cell}} = 1.2 - 0 = 1.2 \ V$.

(C) At equilibrium,the cell potential $E_{\text{cell}} = 0$ and the reaction quotient $Q = K$.
The Nernst equation is given by: $E_{\text{cell}} = E_{\text{cell}}^{\circ} - \frac{0.0592}{n} \log_{10} Q$.
Substituting the equilibrium conditions: $0 = E_{\text{cell}}^{\circ} - \frac{0.0592}{n} \log_{10} K$.
Rearranging this gives: $E_{\text{cell}}^{\circ} = \frac{0.0592}{n} \log_{10} K$.

(B) The cell reaction is: $3 Zn_{(s)} + 2 Cr^{3+}_{(0.1 \ M)} \longrightarrow 3 Zn^{2+}_{(0.1 \ M)} + 2 Cr_{(s)}$
Using the Nernst equation: $E_{cell} = E^{\circ}_{cell} - \frac{0.0592 \ V}{n} \log_{10} \frac{[Zn^{2+}]^3}{[Cr^{3+}]^2}$
Here,$n = 6$ (total electrons transferred).
$E_{cell} = 0.02 \ V - \frac{0.0592 \ V}{6} \log_{10} \frac{(0.1)^3}{(0.1)^2}$
$E_{cell} = 0.02 \ V - \frac{0.0592 \ V}{6} \log_{10} (0.1)$
Since $\log_{10} (0.1) = -1$,we have:
$E_{cell} = 0.02 \ V - \frac{0.0592 \ V}{6} \times (-1)$
$E_{cell} = 0.02 \ V + 0.00987 \ V \approx 0.03 \ V$.

(A) The relationship between standard cell potential and equilibrium constant is given by the formula: $E_{cell}^{\circ} = \frac{0.0592}{n} \log_{10} K$ at $298 \ K$.
The given reaction is $2 Cu_{(aq)}^{+} \longrightarrow Cu_{(aq)}^{2+} + Cu_{(s)}$.
The half-reactions are:
Reduction: $Cu_{(aq)}^{+} + e^{-} \longrightarrow Cu_{(s)}$
Oxidation: $Cu_{(aq)}^{+} \longrightarrow Cu_{(aq)}^{2+} + e^{-}$
Thus,the number of electrons transferred,$n = 1$.
Substituting the values: $E_{cell}^{\circ} = \frac{0.0592}{1} \log_{10} (1.2 \times 10^6)$.
$E_{cell}^{\circ} = 0.0592 (\log 1.2 + \log 10^6)$.
$E_{cell}^{\circ} = 0.0592 (0.079 + 6)$.
$E_{cell}^{\circ} = 0.0592 \times 6.079 \approx 0.36 \ V$.

(A) The cell reaction is as follows:

$Cd_{(s)} \longrightarrow Cd_{(aq)}^{2+} + 2e^{-}$	(Oxidation at anode)
$Cu_{(aq)}^{2+} + 2e^{-} \longrightarrow Cu_{(s)}$	(Reduction at cathode)
$Cd_{(s)} + Cu_{(aq)}^{2+} \longrightarrow Cd_{(aq)}^{2+} + Cu_{(s)}$	(Overall cell reaction)

Using the Nernst equation at $25^{\circ} C$ $(298 \ K)$:
$E_{cell} = E_{cell}^{o} - \frac{0.0592}{n} \log \frac{[Product]}{[Reactant]}$
Here,the number of electrons transferred $n = 2$.
Substituting the concentrations:
$E_{cell} = E_{cell}^{o} - \frac{0.0592}{2} \log \frac{[Cd^{2+}]}{[Cu^{2+}]}$
$E_{cell} = E_{cell}^{o} - 0.0296 \log \frac{[Cd^{2+}]}{[Cu^{2+}]}$

(B) We know the Nernst equation: $E_{cell} = E_{cell}^0 - \frac{0.0592}{n} \log_{10} Q$.
At equilibrium,$E_{cell} = 0$ and $Q = K$.
Substituting these values,we get: $0 = E_{cell}^0 - \frac{0.0592}{n} \log_{10} K$.
Therefore,the correct relation is: $E_{cell}^0 = \frac{0.0592}{n} \log_{10} K$.

(A) The cell reaction is $Zn_{(s)} + Pb^{2+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Pb_{(s)}$.
Using the Nernst equation: $E_{\text{cell}} = E_{\text{cell}}^0 - \frac{0.0591}{n} \log \frac{[Zn^{2+}]}{[Pb^{2+}]}$.
Here,$n = 2$. Initially,$[Zn^{2+}] = 1 \ M$ and $[Pb^{2+}] = 1 \ M$,so $E_{\text{cell, initial}} = E_{\text{cell}}^0$.
When the concentration of $Zn^{2+}$ at the anode increases $10$ times,$[Zn^{2+}] = 10 \ M$.
The new potential is $E_{\text{cell, final}} = E_{\text{cell}}^0 - \frac{0.0591}{2} \log \frac{10}{1} = E_{\text{cell}}^0 - 0.02955 \log(10) = E_{\text{cell}}^0 - 0.0296 \ V$.
Therefore,the potential decreases by $0.0296 \ V$.

(A) The cell reaction is: $2 Ag_{(aq)}^{+} + Cd_{(s)} \longrightarrow 2 Ag_{(s)} + Cd_{(aq)}^{2+}$
Here,the number of electrons transferred,$n = 2$.
The standard free energy change is given by the formula: $\Delta G^{\circ} = -nFE^{\circ}_{cell}$
Substituting the values: $\Delta G^{\circ} = -2 \times 96500 \ C/mol \times 1.20 \ V$
$\Delta G^{\circ} = -231600 \ J/mol = -231.6 \ kJ/mol$

(A) Given: $E^{\circ}_{\text{cell}} = 1.049 \ V$,$n = 2$.
The relationship between $E^{\circ}_{\text{cell}}$ and the equilibrium constant $K$ is given by the Nernst equation at $298 \ K$:
$E^{\circ}_{\text{cell}} = \frac{0.0592}{n} \log_{10} K$
Rearranging for $\log_{10} K$:
$\log_{10} K = \frac{E^{\circ}_{\text{cell}} \times n}{0.0592} = \frac{1.049 \times 2}{0.0592}$
$\log_{10} K = \frac{2.098}{0.0592} \approx 35.439$
Taking the antilog:
$K = 10^{35.439} = 2.75 \times 10^{35}$

(B) The general form of the Nernst equation is:
$E_{cell} = E^o_{cell} - \frac{0.0591}{n} \log \frac{[Product]}{[Reactant]}$
For the given cell reaction:
Oxidation: $Zn_{(s)} \rightarrow Zn^{2+}_{(aq)} (0.01 \ M) + 2e^-$
Reduction: $Cu^{2+}_{(aq)} (1.0 \ M) + 2e^- \rightarrow Cu_{(s)}$
Here,$n = 2$.
Applying the Nernst equation:
$E_{cell} = 2.0 - \frac{0.0591}{2} \log \left( \frac{[Zn^{2+}]}{[Cu^{2+}]} \right)$
$E_{cell} = 2.0 - \frac{0.0591}{2} \log \left( \frac{0.01}{1.0} \right)$
$E_{cell} = 2.0 - \frac{0.0591}{2} \log (10^{-2})$
$E_{cell} = 2.0 - \frac{0.0591}{2} \times (-2)$
$E_{cell} = 2.0 + 0.0591 = 2.0591 \ V \approx 2.0592 \ V$

(A) The general Nernst equation for a half-cell reaction is given by:
$E = E^{o} - \frac{0.0592}{n} \log \frac{[\text{Product}]}{[\text{Reactant}]}$
For the given half-cell reaction:
$Cu^{2+}_{(aq)} + e^{-} \rightarrow Cu^{+}_{(aq)}$
The number of electrons involved is $n = 1$.
Substituting the values into the Nernst equation:
$E_{Cu^{2+}/Cu^{+}} = E^{o}_{Cu^{2+}/Cu^{+}} - \frac{0.0592}{1} \log \frac{[Cu^{+}]}{[Cu^{2+}]}$
Thus,the correct option is $A$.

(D) The reduction reaction for a hydrogen electrode is: $2H^{+} + 2e^{-} \longrightarrow H_{2}(g)$.
According to the Nernst equation at $298 \ K$:
$E = E^{\circ} - \frac{0.0591}{n} \log \frac{P_{H_{2}}}{[H^{+}]^{2}}$.
Given $E^{\circ} = 0 \ V$,$n = 2$,and $P_{H_{2}} = 1 \ atm$:
$E = 0 - \frac{0.0591}{2} \log \frac{1}{[H^{+}]^{2}}$.
$E = -\frac{0.0591}{2} \times (-2 \log [H^{+}])$.
Since $pH = -\log [H^{+}]$,we get:
$E = -0.0591 \ pH$.

(D) The half-cell reaction for a hydrogen electrode is: $2H^+ (aq) + 2e^- \rightarrow H_2 (g)$.
Using the Nernst equation at $298 \ K$: $E = E^0 - \frac{0.0591}{n} \log \frac{1}{[H^+]}$.
Since $E^0 = 0 \ V$ for the standard hydrogen electrode and $n = 1$ for the reduction of $H^+$,the equation becomes: $E = 0 - 0.0591 \times pH$.
Given $pH = 10$,we have: $E = -0.0591 \times 10 = -0.591 \ V$.
Rounding to two decimal places,the potential is $-0.59 \ V$.

(D) The cell reaction is as follows:
Anode (Oxidation): $Mg_{(s)} \rightarrow Mg_{(aq)}^{2+} + 2e^-$
Cathode (Reduction): $Cl_{2_{(g)}} + 2e^- \rightarrow 2Cl_{(aq)}^{-}$
Overall reaction: $Mg_{(s)} + Cl_{2_{(g)}} \rightarrow Mg_{(aq)}^{2+} + 2Cl_{(aq)}^{-}$
The Nernst equation is given by: $E_{cell} = E_{cell}^0 - \frac{0.059}{n} \log Q$
Here,$n = 2$ and the reaction quotient $Q = \frac{[Mg^{2+}][Cl^{-}]^2}{P_{Cl_2}}$.
Since $P_{Cl_2} = 1 \ bar$,$Q = [Mg^{2+}][Cl^{-}]^2$.
Thus,the correct equation is $E_{cell} = E_{cell}^0 - \frac{0.059}{2} \log [Mg^{2+}][Cl^{-}]^2$.

(B) The cell reaction is: $Zn_{(s)} + Cu_{(0.02 \ M)}^{2+} \rightarrow Zn_{(x \ M)}^{2+} + Cu_{(s)}$.
According to the Nernst equation: $E_{cell} = E_{cell}^0 - \frac{0.059}{2} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$.
Substituting the values: $E_{cell} = E_{cell}^0 - 0.0295 \log \frac{x}{0.02}$.
To increase $E_{cell}$,the term $\log \frac{x}{0.02}$ must be negative.
This occurs when $\frac{x}{0.02} < 1$,which implies $x < 0.02 \ M$.

(A) The oxidation half-reaction for the hydrogen electrode is:
$H_{2(g)} \rightarrow 2H^+_{(aq)} + 2e^-$
Using the Nernst equation for oxidation potential:
$E_{ox} = E^{\circ}_{ox} - \frac{0.0591}{n} \log Q$
For the standard hydrogen electrode,$E^{\circ}_{ox} = 0.00 \ V$ and $n = 2$.
The reaction quotient $Q$ is given by $[H^+]^2 / P_{H_2}$.
Given $pH = 3$,so $[H^+] = 10^{-3} \ M$.
$Q = \frac{(10^{-3})^2}{1} = 10^{-6}$.
$E_{ox} = 0 - \frac{0.0591}{2} \log(10^{-6})$
$E_{ox} = -0.02955 \times (-6) = 0.1773 \ V$.
Thus,the oxidation potential is approximately $0.177 \ V$.

(A) The oxidation reaction for the hydrogen electrode is: $H_2(g) \rightarrow 2H^+(aq) + 2e^-$.
Using the Nernst equation for the oxidation potential: $E_{ox} = E^0_{ox} - \frac{0.0591}{n} \log [H^+]^2$.
Since $E^0_{ox}$ for hydrogen is $0 \ V$ and $n = 2$,we have: $0.177 = 0 - \frac{0.0591}{2} \times 2 \log [H^+]$.
$0.177 = -0.0591 \times \log [H^+]$.
Since $pH = -\log [H^+]$,we get: $0.177 = 0.0591 \times pH$.
$pH = \frac{0.177}{0.0591} = 3$.
Therefore,the correct option is $A$.

(A) The cell reaction is: $2Al_{(s)} + 3Zn_{(aq)}^{2+} \rightarrow 2Al_{(aq)}^{3+} + 3Zn_{(s)}$.
Here,the number of electrons involved in the reaction is $n = 6$.
The Nernst equation is given by: $E_{cell} = E_{cell}^0 - \frac{0.059}{n} \log \frac{[Product]}{[Reactant]}$.
Substituting the values: $E_{cell} = E_{cell}^0 - \frac{0.059}{6} \log \frac{[Al^{3+}]^2}{[Zn^{2+}]^3}$.
Thus,option $A$ is correct.

(B) The reduction reaction for a hydrogen half-cell is: $2H^{+} (aq) + 2e^{-} \rightarrow H_2 (g)$.
Using the Nernst equation: $E_{red} = E^{\circ}_{red} - \frac{0.0591}{n} \log \frac{P_{H_2}}{[H^{+}]^2}$.
Since $E^{\circ}_{red} = 0 \ V$ and $n = 2$,the equation becomes: $E_{red} = -0.02955 \log \frac{P_{H_2}}{[H^{+}]^2}$.
For $E_{red}$ to be negative,$\log \frac{P_{H_2}}{[H^{+}]^2}$ must be positive,which means $\frac{P_{H_2}}{[H^{+}]^2} > 1$.
Checking the options:
$A$: $\frac{2}{2^2} = 0.5 < 1$ (Positive potential).
$B$: $\frac{2}{1^2} = 2 > 1$ (Negative potential).
$C$: $\frac{1}{1^2} = 1$ $(E_{red} = 0 \ V)$.
$D$: $\frac{1}{2^2} = 0.25 < 1$ (Positive potential).
Therefore,option $B$ is correct.

(B) The potential of a hydrogen electrode is given by the Nernst equation: $E_{H^{+}/H_2} = E^{\circ}_{H^{+}/H_2} - \frac{0.059}{n} \log \frac{1}{[H^{+}]}$.
Given $pH = 1$,we know that $-\log [H^{+}] = 1$,which implies $[H^{+}] = 10^{-1} \ M$.
For a standard hydrogen electrode,$E^{\circ}_{H^{+}/H_2} = 0.0 \ V$ and $n = 1$.
Substituting the values: $E = 0.0 - \frac{0.059}{1} \log \frac{1}{10^{-1}}$.
$E = -0.059 \times \log(10) = -0.059 \times 1 = -0.059 \ V$.

(B) The relationship between standard cell potential and equilibrium constant is given by the formula:
$E_{cell}^0 = \frac{0.0591}{n} \log K_c$
Here,$n = 2$ (number of electrons transferred).
$0.46 = \frac{0.0591}{2} \log K_c$
$\log K_c = \frac{0.46 \times 2}{0.0591} = 15.5668$
$K_c = \text{antilog}(15.5668) \approx 3.92 \times 10^{15}$

(D) The relationship between standard $emf$ $(E_{\text{cell}}^{\circ})$ and equilibrium constant $(K_{c})$ is given by the Nernst equation at $298 \ K$:
$E_{\text{cell}}^{\circ} = \frac{0.0591}{n} \log K_{c}$
Given $E_{\text{cell}}^{\circ} = 0.59 \ V$ and $n = 3$:
$0.59 = \frac{0.0591}{3} \log K_{c}$
Approximating $0.0591 \approx 0.059$:
$0.59 = \frac{0.059}{3} \log K_{c}$
$\log K_{c} = \frac{0.59 \times 3}{0.059} = 10 \times 3 = 30$
$K_{c} = 10^{30}$

(C) The Nernst equation for the given half-cell reaction is:
$E = E^{\circ} - \frac{0.0591}{6} \log \frac{[Cr^{3+}]^{2}}{[Cr_{2}O_{7}^{2-}][H^{+}]^{14}}$
Substituting the given values:
$1.067 = 1.33 - \frac{0.0591}{6} \log \frac{(15 \times 10^{-3})^{2}}{(4.5 \times 10^{-3})[H^{+}]^{14}}$
$1.067 - 1.33 = -\frac{0.0591}{6} \log \frac{225 \times 10^{-6}}{4.5 \times 10^{-3} \cdot [H^{+}]^{14}}$
$-0.263 = -0.00985 \log \frac{0.05}{[H^{+}]^{14}}$
$26.7 = \log 0.05 + 14 pH$
$26.7 = -1.3 + 14 pH$
$28 = 14 pH$
$pH = 2$