For a Daniell cell,$E^0_{cell} = 1.1 \ V$. How is $K_c$ represented for the reaction occurring in the Daniell cell?

  • A
    $K_c = 10^{2.2/0.059}$
  • B
    $K_c = 10^{-0.059/1.1}$
  • C
    $K_c = 10^{-2.2/0.059}$
  • D
    $K_c = 10^{0.059/1.1}$

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Similar Questions

In the given electrochemical cell,$Ag_{(s)} | AgCl_{(s)} | Cl^-_{(aq)}, Fe^{2+}_{(aq)}, Fe^{3+}_{(aq)} | Pt_{(s)}$ at $298 \ K$,the cell potential $(E_{cell})$ will increase when :
$(A)$ Concentration of $Fe^{2+}$ is increased.
$(B)$ Concentration of $Fe^{3+}$ is decreased.
$(C)$ Concentration of $Fe^{2+}$ is decreased.
$(D)$ Concentration of $Fe^{3+}$ is increased.
$(E)$ Concentration of $Cl^-$ is increased.
Choose the correct answer from the options given below :

Calculate $E_{cell}^{\circ}$ if the equilibrium constant for the following reaction is $1.2 \times 10^6$.
$2 Cu_{(aq)}^{+} \longrightarrow Cu_{(aq)}^{2+} + Cu_{(s)}$ (in $V$)

The Nernst equation is related to:

Write the Nernst equation for the $E_{cell}$ reaction in the Daniell cell. How will the $E_{cell}$ be affected when the concentration of $Zn^{2+}$ ions is increased?

What is $E_{cell}$ (in $V$) of the following cell at $298 \ K$ ?
$(E^{\ominus}_{Zn^{2+}/Zn} = -0.76 \ V ; E^{\ominus}_{Ni^{2+}/Ni} = -0.25 \ V ; \frac{2.303 RT}{F} = 0.06 \ V)$
$Zn_{(s)} | Zn^{2+} (0.01 \ M) || Ni^{2+} (0.1 \ M) | Ni_{(s)}$

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