Write a note on the relation between Gibbs free energy and cell potential for a cell reaction.
(N/A) The electrical work $(\omega_{\text{ele}})$ done in one second is equal to the electrical potential multiplied by the total charge passed: $\omega_{\text{ele}} = E \times Q$.
To obtain maximum work from a galvanic cell, the charge must be passed reversibly. The reversible work done by a galvanic cell is equal to the decrease in its Gibbs energy: $\omega_{\text{ele}} = -\Delta_{r}G$.
If the electromotive force $(EMF)$ of the cell is $E_{\text{cell}}$ and the total charge passed is $nF$ (where $n$ is the number of moles of electrons transferred and $F$ is Faraday's constant), then the Gibbs energy change for the reaction is given by: $\Delta_{r}G = -nF E_{\text{cell}}$.
Here, $E_{\text{cell}}$ is an intensive property, while $\Delta_{r}G$ is an extensive thermodynamic property, and its value depends on the value of $n$.
If all species taking part in the reaction are at unit activity, then $E_{\text{cell}} = E_{\text{cell}}^{o}$ and $\Delta_{r}G^{o} = -nF E_{\text{cell}}^{o}$.
Thus, the standard Gibbs energy change $\Delta_{r}G^{o}$ can be determined by measuring the standard cell potential $E_{\text{cell}}^{o}$.
Using the value of $\Delta_{r}G^{o}$, the equilibrium constant $(K)$ can be calculated using the formula: $\Delta_{r}G^{o} = -RT \ln K = -2.303 RT \log K$.
To obtain maximum work from a galvanic cell, the charge must be passed reversibly. The reversible work done by a galvanic cell is equal to the decrease in its Gibbs energy: $\omega_{\text{ele}} = -\Delta_{r}G$.
If the electromotive force $(EMF)$ of the cell is $E_{\text{cell}}$ and the total charge passed is $nF$ (where $n$ is the number of moles of electrons transferred and $F$ is Faraday's constant), then the Gibbs energy change for the reaction is given by: $\Delta_{r}G = -nF E_{\text{cell}}$.
Here, $E_{\text{cell}}$ is an intensive property, while $\Delta_{r}G$ is an extensive thermodynamic property, and its value depends on the value of $n$.
If all species taking part in the reaction are at unit activity, then $E_{\text{cell}} = E_{\text{cell}}^{o}$ and $\Delta_{r}G^{o} = -nF E_{\text{cell}}^{o}$.
Thus, the standard Gibbs energy change $\Delta_{r}G^{o}$ can be determined by measuring the standard cell potential $E_{\text{cell}}^{o}$.
Using the value of $\Delta_{r}G^{o}$, the equilibrium constant $(K)$ can be calculated using the formula: $\Delta_{r}G^{o} = -RT \ln K = -2.303 RT \log K$.
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Similar Questions
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 $2$.
By how much will the potential of the half-cell $Cu^{2+}/Cu$ change if the solution is diluted to $100$ times at $298 \ K$?
Medium
View SolutionFor the cell reaction $Sn_{(s)} + 2Ag^+_{(aq)} \rightarrow Sn^{2+}_{(aq)} + 2Ag_{(s)}$,what happens when the cell voltage increases?
Easy
View SolutionWhich of the following statements is true regarding the cell emf at $298 \ K$ for the cell $Ni_{(s)} | Ni^{2+}(0.01 \ M) || Ag^{+}(0.01 \ M) | Ag_{(s)}$?
In which of the following conditions will the reduction potential of a hydrogen half-cell be negative?
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