For a cell reaction involving a two-electron change,the standard $EMF$ of the cell is $0.295 \ V$ at $25 \ ^\circ C$. The equilibrium constant of the reaction at $25 \ ^\circ C$ will be:

  • A
    $29.5 \times 10^{-2}$
  • B
    $10$
  • C
    $1 \times 10^{10}$
  • D
    $2.95 \times 10^{-10}$

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Similar Questions

Which of the following expressions is used to find the cell potential of the $Cd_{(s)} | Cd_{(aq)}^{2+} || Cu_{(aq)}^{2+} | Cu_{(s)}$ cell at $25^{\circ} C$?

Calculate the potential of a hydrogen electrode in contact with a solution whose $pH$ is $10$. (in $V$)

For $Cr_2O_7^{2-} + 14H^{+} + 6e^- \longrightarrow 2Cr^{3+} + 7H_2O$,$E^0 = 1.33 \ V$. Given $[Cr_2O_7^{2-}] = 4.5 \ mmol$,$[Cr^{3+}] = 1.5 \ mmol$ and $E = 1.067 \ V$,calculate the $pH$ of the solution.

If a cell has a standard electrode potential of $0.295 \ V$ and $n = 2$,calculate its equilibrium constant at $298 \ K$.

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 $[Cu^{2+}] = C_1$.

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