If $A$ is the reactant and $P$ is the product,which one of the following is the correct form of the Nernst equation?
- A$\frac{[A]}{[P]}=\exp \left(\frac{R T}{n F}(E-E^{\circ})\right)$
- B$\frac{[A]}{[P]}=\exp \left(\frac{n F}{R T}(E-E^{\circ})\right)$
- C$\frac{[A]}{[P]}=\exp \left(-\frac{n F}{R T}(E-E^{\circ})\right)$
- D$E=E^{\circ}-\frac{R T}{n F} \ln \frac{[A]}{[P]}$
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Similar Questions
Under which of the following conditions is the $E$ value of the cell for the given reaction maximum?
$Zn_{(s)} + Cu^{2+}_{(aq)} \rightleftharpoons Cu_{(s)} + Zn^{2+}_{(aq)}$
$\left( \frac{2.303 RT}{F} \text{ at } 298 \ K = 0.059 \ V, E^{\circ}_{Zn^{2+}/Zn} = -0.76 \ V, E^{\circ}_{Cu^{2+}/Cu} = +0.34 \ V \right)$
Let $[Zn^{2+}] = C_2$ and $[Cu^{2+}] = C_1$.
$Zn_{(s)} + Cu^{2+}_{(aq)} \rightleftharpoons Cu_{(s)} + Zn^{2+}_{(aq)}$
$\left( \frac{2.303 RT}{F} \text{ at } 298 \ K = 0.059 \ V, E^{\circ}_{Zn^{2+}/Zn} = -0.76 \ V, E^{\circ}_{Cu^{2+}/Cu} = +0.34 \ V \right)$
Let $[Zn^{2+}] = C_2$ and $[Cu^{2+}] = C_1$.
What will be the reduction potential of $Cu$ in an aqueous solution with $pH = 12$? Given that the $K_{sp}$ of $Cu(OH)_2$ is $1 \times 10^{-19}$ and $E^o_{Cu^{2+}/Cu} = 0.34 \ V$.
Medium
View SolutionFor the cell $Cu_{(s)}|Cu^{2+}_{(aq)}(0.1 \ M) || Ag^{+}_{(aq)}(0.01 \ M)| Ag_{(s)}$,the cell potential $E_{1} = 0.3095 \ V$. For the cell $Cu_{(s)}|Cu^{2+}_{(aq)}(0.01 \ M) || Ag^{+}_{(aq)}(0.001 \ M)| Ag_{(s)}$,the cell potential $= ..... \times 10^{-2} \ V$. (Round off to the Nearest Integer). [Use: $\frac{2.303 \ RT}{F} = 0.059$]
DifficultJEE MAIN 2021
View SolutionWhat is $E_{cell}$ (in $V$) of the following cell at $298 \ K$ ?
$(E^{\ominus}_{Zn^{2+}/Zn} = -0.76 \ V ; E^{\ominus}_{Ni^{2+}/Ni} = -0.25 \ V ; \frac{2.303 RT}{F} = 0.06 \ V)$
$Zn_{(s)} | Zn^{2+} (0.01 \ M) || Ni^{2+} (0.1 \ M) | Ni_{(s)}$
$(E^{\ominus}_{Zn^{2+}/Zn} = -0.76 \ V ; E^{\ominus}_{Ni^{2+}/Ni} = -0.25 \ V ; \frac{2.303 RT}{F} = 0.06 \ V)$
$Zn_{(s)} | Zn^{2+} (0.01 \ M) || Ni^{2+} (0.1 \ M) | Ni_{(s)}$
For the cell $Zn_{(s)} | Zn^{2+} (0.1 \ M) || Fe^{2+} (0.01 \ M) | Fe_{(s)}$ at $298 \ K$,$E_{cell} = 0.2905 \ V$. What is the equilibrium constant $(K_c)$ for the reaction $Zn_{(s)} + Fe^{2+}_{(aq)} \rightleftharpoons Zn^{2+}_{(aq)} + Fe_{(s)}$?
Difficult
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