Nernst equation and ECS Questions in English

(B) The reduction half-reaction for the copper electrode is: $Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$.
Using the Nernst equation at $25^{\circ} C$ $(298 \ K)$:
$E_{Cu^{2+}/Cu} = E^{\circ}_{Cu^{2+}/Cu} - \frac{0.0591}{n} \log \frac{1}{[Cu^{2+}]}$
Given $E^{\circ} = +0.34 \ V$,$n = 2$,and $[Cu^{2+}] = 0.01 \ M = 10^{-2} \ M$.
$E_{Cu^{2+}/Cu} = 0.34 - \frac{0.0591}{2} \log \frac{1}{10^{-2}}$
$E_{Cu^{2+}/Cu} = 0.34 - \frac{0.0591}{2} \log(10^2)$
$E_{Cu^{2+}/Cu} = 0.34 - \frac{0.0591}{2} \times 2$
$E_{Cu^{2+}/Cu} = 0.34 - 0.0591 = 0.2809 \ V$.

(C) The reduction reaction for a hydrogen half-cell is: $2H^{+} + 2e^{-} \rightarrow H_{2}$
Here,$n = 2$ and the reaction quotient $Q = \frac{p(H_{2})}{[H^{+}]^{2}}$.
Using the Nernst equation: $E_{H^{+}/H_{2}} = E^{0}_{H^{+}/H_{2}} - \frac{0.059}{n} \log Q$.
Since $E^{0}_{H^{+}/H_{2}} = 0 \ V$,we have $E_{H^{+}/H_{2}} = -\frac{0.059}{2} \log Q$.
For $E_{H^{+}/H_{2}}$ to be negative,$\log Q$ must be positive,which means $Q > 1$.
Evaluating the options:
$(A)$ $Q = \frac{1}{1^{2}} = 1$ (Not negative)
$(B)$ $Q = \frac{1}{2^{2}} = 0.25 < 1$ (Positive potential)
$(C)$ $Q = \frac{2}{1^{2}} = 2 > 1$ (Negative potential)
$(D)$ $Q = \frac{2}{2^{2}} = 0.5 < 1$ (Positive potential)
Therefore,the correct option is $(C)$.

(A) The cell reaction is: $2Ag_{(s)} + Zn^{2+}(0.1 \ M) \longrightarrow 2Ag^{+}(0.1 \ M) + Zn_{(s)}$
Using the Nernst equation at $298 \ K$:
$E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log \frac{[Ag^{+}]^{2}}{[Zn^{2+}]}$
Here,$n = 2$,$[Ag^{+}] = 0.1 \ M$,and $[Zn^{2+}] = 0.1 \ M$.
$E_{cell} = -1.562 - \frac{0.0591}{2} \log \frac{(0.1)^{2}}{0.1}$
$E_{cell} = -1.562 - 0.02955 \log(0.1)$
$E_{cell} = -1.562 - 0.02955 \times (-1)$
$E_{cell} = -1.562 + 0.02955 = -1.53245 \ V \approx -1.532 \ V$

(C) The half-cell reaction for a hydrogen electrode is: $2H^+ (aq) + 2e^- \rightarrow H_2 (g)$.
Using the Nernst equation: $E = E^0 - \frac{0.0591}{n} \log \frac{P_{H_2}}{[H^+]^2}$.
Given $E^0 = 0 \ V$,$n = 2$,$P_{H_2} = 1 \ atm$,and $[H^+] = 10^{-pH} = 10^{-10} \ M$.
Substituting the values: $E = 0 - \frac{0.0591}{2} \log \frac{1}{(10^{-10})^2}$.
$E = -\frac{0.0591}{2} \log (10^{20})$.
$E = -\frac{0.0591}{2} \times 20$.
$E = -0.0591 \times 10 = -0.591 \ V$.
Thus,the potential is approximately $-0.59 \ V$.

(A, B, D) : In $CrO_5$ (butterfly structure),there are four peroxide linkages and one double bond. The oxidation state of $Cr$ is $+6$. This statement is correct.
$B$: For the reaction $N_2O_{4(g)} \rightarrow 2NO_{2(g)}$,$\Delta n_g = 2 - 1 = 1$. Since $\Delta H = \Delta U + \Delta n_g RT$,and $\Delta n_g > 0$,$\Delta H > \Delta U$. This statement is correct.
$C$: For $0.1 \ N \ H_2SO_4$,$[H^+] = 0.1 \ N$. For $0.1 \ N \ HCl$,$[H^+] = 0.1 \ N$. Since both have the same $[H^+]$,their pH is the same. This statement is incorrect.
$D$: The value of $\frac{2.303 RT}{F}$ at $25^{\circ} C$ $(298 \ K)$ is $\frac{2.303 \times 8.314 \times 298}{96500} \approx 0.0591 \ V$. This statement is correct.

(C) For a concentration cell,the cell reaction is $M^{+}(c_{cathode}) \rightarrow M^{+}(c_{anode})$.
Using the Nernst equation: $E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log \frac{[M^{+}]_{anode}}{[M^{+}]_{cathode}}$.
Since $E^{\circ}_{cell} = 0$ for a concentration cell,$E_{cell} = -0.0591 \log \frac{c_{anode}}{c_{cathode}} = 0.0591 \log \frac{c_{cathode}}{c_{anode}}$.
For $E_{cell} > 0$,we must have $\log \frac{c_{cathode}}{c_{anode}} > 0$,which implies $c_{cathode} > c_{anode}$.
Case $I$: If $c_{1}$ is at the cathode,then $c_{1} > c_{2}$ (Option $C$ is correct).
Case $II$: If $c_{1}$ is at the anode,then $c_{2} > c_{1}$ (Option $D$ is incorrect).

(D) The cell reaction is: $Ag_{(s)} + Cl^-_{(aq)} + Fe^{3+}_{(aq)} \rightarrow AgCl_{(s)} + Fe^{2+}_{(aq)}$.
According to the Nernst equation: $E_{cell} = E^o_{cell} - \frac{0.0591}{1} \log \frac{[Fe^{2+}]}{[Cl^-][Fe^{3+}]}$.
To increase $E_{cell}$,the value of the logarithmic term $\frac{[Fe^{2+}]}{[Cl^-][Fe^{3+}]}$ must decrease.
This occurs when $[Fe^{2+}]$ is decreased,$[Fe^{3+}]$ is increased,or $[Cl^-]$ is increased.
Therefore,the correct options are $C, D,$ and $E$.

(B) The cell reaction is $M^{2+} + H_2O \rightarrow M + \frac{1}{2}O_2 + 2H^+$.
For the reaction to be spontaneous,$E_{cell} > 0$.
At the limiting condition,$E_{cell} = 0$,so $E_{cathode} = E_{anode}$.
$E_{cathode} = E^{\circ}_{M^{2+}/M} - \frac{0.059}{2} \log \frac{1}{[M^{2+}]} = 0.994 - 0 = 0.994 \ V$.
$E_{anode} = E^{\circ}_{O_2/H_2O} - \frac{0.059}{4} \log \frac{1}{[H^+]^4 P_{O_2}^{1/2}} = 1.23 + \frac{0.059}{4} \log ([H^+]^4 \times 1) = 1.23 + 0.059 \log [H^+] = 1.23 - 0.059 \times pH$.
Equating $E_{cathode} = E_{anode}$:
$0.994 = 1.23 - 0.059 \times pH$.
$0.059 \times pH = 1.23 - 0.994 = 0.236$.
$pH = \frac{0.236}{0.059} = 4$.
Thus,the nearest integer is $4$.

(C) The cell reaction is: $3HSnO_2^- + Bi_2O_3 + 3H_2O + 6OH^- \rightarrow 3Sn(OH)_6^{2-} + 2Bi$.
The number of electrons involved in the balanced equation is $n = 6$.
The standard cell potential is $E^0_{\text{cell}} = E^0_{\text{cathode}} - E^0_{\text{anode}} = -0.44 - (-0.90) = +0.46 \ V$.
Applying the Nernst equation at $298 \ K$: $E_{\text{cell}} = E^0_{\text{cell}} - \frac{0.0591}{n} \log Q$.
Using $0.06$ as an approximation for $\frac{0.0591}{1}$: $E_{\text{cell}} = 0.46 - \frac{0.06}{6} \log(10^6)$.
$E_{\text{cell}} = 0.46 - 0.01 \times 6 = 0.46 - 0.06 = 0.40 \ V$.
Thus,$E_{\text{cell}} = 4 \times 10^{-1} \ V$.
The value is $4$.

(A) The relationship between the standard cell potential $(E^0_{cell})$ and the equilibrium constant $(K_c)$ is given by the Nernst equation at equilibrium: $E^0_{cell} = \frac{0.059}{n} \log K_c$ at $298 \ K$.
For a Daniell cell,the cell reaction is $Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$,where the number of electrons transferred is $n = 2$.
Substituting the values into the formula: $1.1 = \frac{0.059}{2} \log K_c$.
Multiplying both sides by $2$,we get $2.2 = 0.059 \log K_c$.
Therefore,$\log K_c = \frac{2.2}{0.059}$.
Converting from logarithmic form to exponential form,we get $K_c = 10^{2.2/0.059}$.
Thus,option $(A)$ is correct.

(D) The relationship between the standard cell potential $(E^\circ_{\text{cell}})$ and the equilibrium constant $(K)$ is given by the Nernst equation at equilibrium:
$\log K = \frac{n E^\circ_{\text{cell}}}{0.0591}$
Here,the number of electrons transferred $(n)$ is $2$,and $E^\circ_{\text{cell}} = 0.46 \text{ V}$.
Substituting the values:
$\log K = \frac{2 \times 0.46}{0.0591} = \frac{0.92}{0.0591} \approx 15.5668$
Now,$K = 10^{15.5668} = 10^{0.5668} \times 10^{15} \approx 3.68 \times 10^{15}$.
Comparing this with the given options,the closest value is $3.92 \times 10^{15}$.

(B) $1$. The cell reaction is: $Zn(s) + 2Ag^+(aq) \to Zn^{2+}(aq) + 2Ag(s)$.
$2$. Calculate the standard cell potential: $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.80 \text{ V} - (-0.76 \text{ V}) = 1.56 \text{ V}$.
$3$. Apply the Nernst equation at $298 \text{ K}$: $E_{cell} = E^\circ_{cell} - \frac{0.059}{n} \log \frac{[Zn^{2+}]}{[Ag^+]^2}$.
$4$. Here,$n = 2$,$[Zn^{2+}] = 1 \text{ M}$,and $[Ag^+] = x$. Substituting the values: $1.60 = 1.56 - \frac{0.059}{2} \log \frac{1}{x^2}$.
$5$. Simplify the equation: $1.60 = 1.56 - 0.0295 \times (-2 \log x) = 1.56 + 0.059 \log x$.
$6$. Solve for $\log x$: $0.04 = 0.059 \log x$,which gives $\log x = \frac{0.04}{0.059} = \frac{4}{5.9}$.

(B) The cell reaction involves the oxidation of $M$ and reduction of $Fe^{3+}$.
$M \rightarrow M^{x+} + xe^-$ (Anode)
$Fe^{3+} + 3e^- \rightarrow Fe$ (Cathode)
Balanced cell reaction: $3M + xFe^{3+} \rightarrow 3M^{x+} + xFe$.
The number of electrons transferred is $n = 3x$.
Standard cell potential: $E^{\ominus}_{cell} = E^{\ominus}_{cathode} - E^{\ominus}_{anode} = 0.15 - (-0.036) = 0.186 \text{ V}$.
Using the Nernst equation: $E_{cell} = E^{\ominus}_{cell} - \frac{0.059}{n} \log Q$.
$0.2057 = 0.186 - \frac{0.059}{3x} \log(10^{-2})$.
$0.2057 - 0.186 = -\frac{0.059}{3x} \times (-2)$.
$0.0197 = \frac{0.118}{3x}$.
$3x = \frac{0.118}{0.0197} \approx 6$.
$x = 2$.

(B) The half-cell reaction is: $H_2(g) \to 2H^+(aq) + 2e^-$.
Using the Nernst equation for the oxidation potential: $E = E^\circ - \frac{0.059}{n} \log Q$.
Here,$n = 2$,$E^\circ = 0 \text{ V}$,$[H^+] = 0.02 \text{ M}$,and $P_{H_2} = 2 \text{ atm}$.
$Q = \frac{[H^+]^2}{P_{H_2}} = \frac{(0.02)^2}{2} = \frac{0.0004}{2} = 0.0002 = 2 \times 10^{-4}$.
$E = 0 - \frac{0.059}{2} \log(2 \times 10^{-4})$.
$E = -0.0295 \times (\log 2 + \log 10^{-4}) = -0.0295 \times (0.3010 - 4) = -0.0295 \times (-3.699) \approx 0.109 \text{ V}$.