Explain the equilibrium state in a Daniell cell and derive its equilibrium constant. Given: $E^o_{cell} = 1.1 \ V$.
(N/A) The Daniell cell reaction is: $Zn_{(s)} + Cu_{(aq)}^{2+} \rightarrow Zn_{(aq)}^{2+} + Cu_{(s)}$.
At the anode,$Zn$ is oxidized to $Zn^{2+}$,increasing its concentration,while at the cathode,$Cu^{2+}$ is reduced to $Cu$,decreasing its concentration. Consequently,the cell potential decreases over time.
When the concentrations of $Zn^{2+}$ and $Cu^{2+}$ ions no longer change and the voltmeter reads $0 \ V$,the system has reached equilibrium.
At equilibrium: $Zn_{(s)} + Cu_{(aq)}^{2+} \rightleftharpoons Zn_{(aq)}^{2+} + Cu_{(s)}$.
The equilibrium constant is $K_C = \frac{[Zn^{2+}]}{[Cu^{2+}]}$.
Using the Nernst equation at equilibrium: $E_{cell} = E^o_{cell} - \frac{0.0591}{n} \log K_C = 0$.
Substituting $E^o_{cell} = 1.1 \ V$ and $n = 2$: $1.1 = \frac{0.0591}{2} \log K_C$.
$\log K_C = \frac{1.1 \times 2}{0.0591} \approx 37.22$.
$K_C = 10^{37.22} \approx 1.64 \times 10^{37}$.
At the anode,$Zn$ is oxidized to $Zn^{2+}$,increasing its concentration,while at the cathode,$Cu^{2+}$ is reduced to $Cu$,decreasing its concentration. Consequently,the cell potential decreases over time.
When the concentrations of $Zn^{2+}$ and $Cu^{2+}$ ions no longer change and the voltmeter reads $0 \ V$,the system has reached equilibrium.
At equilibrium: $Zn_{(s)} + Cu_{(aq)}^{2+} \rightleftharpoons Zn_{(aq)}^{2+} + Cu_{(s)}$.
The equilibrium constant is $K_C = \frac{[Zn^{2+}]}{[Cu^{2+}]}$.
Using the Nernst equation at equilibrium: $E_{cell} = E^o_{cell} - \frac{0.0591}{n} \log K_C = 0$.
Substituting $E^o_{cell} = 1.1 \ V$ and $n = 2$: $1.1 = \frac{0.0591}{2} \log K_C$.
$\log K_C = \frac{1.1 \times 2}{0.0591} \approx 37.22$.
$K_C = 10^{37.22} \approx 1.64 \times 10^{37}$.
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$E_{M^{3+} / M}^0 = 0.6 \ V$ and $E_{N^{2+} / N}^0 = 0.1 \ V$ (in $V$)
The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. $A$ simple model for such a concentration cell involving a metal $M$ is:
$M_{(s)} \mid M^{+}(aq; 0.05 \ M) \parallel M^{+}(aq; 1 \ M) \mid M_{(s)}$
For the above electrolytic cell the magnitude of the cell potential $|E_{cell}|=70 \ mV$.
$1.$ For the above cell
$(A)$ $E_{cell} < 0 ; \Delta G > 0$ $(B)$ $E_{cell} > 0 ; \Delta G < 0$
$(C)$ $E_{cell} < 0 ; \Delta G^{\circ} > 0$ $(D)$ $E_{cell} > 0 ; \Delta G^{\circ} > 0$
$2.$ If the $0.05 \ M$ solution of $M^{+}$ is replaced by $0.0025 \ M$ $M^{+}$ solution,then the magnitude of the cell potential would be
$(A)$ $35 \ mV$ $(B)$ $70 \ mV$ $(C)$ $140 \ mV$ $(D)$ $700 \ mV$
Give the answer for questions $1$ and $2$.
$M_{(s)} \mid M^{+}(aq; 0.05 \ M) \parallel M^{+}(aq; 1 \ M) \mid M_{(s)}$
For the above electrolytic cell the magnitude of the cell potential $|E_{cell}|=70 \ mV$.
$1.$ For the above cell
$(A)$ $E_{cell} < 0 ; \Delta G > 0$ $(B)$ $E_{cell} > 0 ; \Delta G < 0$
$(C)$ $E_{cell} < 0 ; \Delta G^{\circ} > 0$ $(D)$ $E_{cell} > 0 ; \Delta G^{\circ} > 0$
$2.$ If the $0.05 \ M$ solution of $M^{+}$ is replaced by $0.0025 \ M$ $M^{+}$ solution,then the magnitude of the cell potential would be
$(A)$ $35 \ mV$ $(B)$ $70 \ mV$ $(C)$ $140 \ mV$ $(D)$ $700 \ mV$
Give the answer for questions $1$ and $2$.
For an electrochemical cell
$Sn_{(s)} | Sn^{2+}(aq, 1 \ M) || Pb^{2+}(aq, 1 \ M) | Pb_{(s)}$
the ratio $\frac{[Sn^{2+}]}{[Pb^{2+}]}$ when this cell attains equilibrium is
(Given $E^{0}_{Sn^{2+}/Sn} = -0.14 \ V$,$E^{0}_{Pb^{2+}/Pb} = -0.13 \ V$,$\frac{2.303 \ RT}{F} = 0.06$)
$Sn_{(s)} | Sn^{2+}(aq, 1 \ M) || Pb^{2+}(aq, 1 \ M) | Pb_{(s)}$
the ratio $\frac{[Sn^{2+}]}{[Pb^{2+}]}$ when this cell attains equilibrium is
(Given $E^{0}_{Sn^{2+}/Sn} = -0.14 \ V$,$E^{0}_{Pb^{2+}/Pb} = -0.13 \ V$,$\frac{2.303 \ RT}{F} = 0.06$)
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