The standard cell potential $E^o$ for the reaction $aA + bB \to cC + dD$ is related to the equilibrium constant $K_c$ by the expression:

  • A
    $E^o = \frac{RT}{nF} \ln K_c$
  • B
    $E^o = -\frac{RT}{nF} \ln K_c$
  • C
    $E^o = \frac{nF}{RT} \ln K_c$
  • D
    $E^o = -\frac{nF}{RT} \ln K_c$

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Similar Questions

For the redox reaction occurring in a cell: $Zn_{(s)} + Cu^{2+}(0.1 \ M) \to Zn^{2+}(1 \ M) + Cu_{(s)}$,if $E^o_{cell} = 1.10 \ V$,calculate the value of $E_{cell}$ in $V$. (Given: $2.303 \frac{RT}{F} = 0.0591$)

The logarithm of the equilibrium constant for the reaction $Pd^{2+}{(aq)} + 4Cl^{-}{(aq)} \rightleftharpoons PdCl_4^{2-}{(aq)}$ is (Nearest integer).
Given: $\frac{2.303 RT}{F} = 0.06 \ V$
$Pd^{2+}{(aq)} + 2e^{-} \rightleftharpoons Pd_{(s)} \quad E^{\circ} = 0.83 \ V$
$PdCl_4^{2-}{(aq)} + 2e^{-} \rightleftharpoons Pd_{(s)} + 4Cl^{-}{(aq)} \quad E^{\circ} = 0.65 \ V$

$A$ solution containing $4.5 \ mM$ of $MnO_4^{-}$ and $15 \ mM$ of $Mn^{2+}$ shows $pH$ of $2$. The potential of the half-cell reaction is $......$. (Given: $\log 15 = 1.176$,$\log 4.5 = 0.653$,and standard potential of $MnO_4^{-} \longrightarrow Mn^{2+}$ is $1.51 \ V$) (in $V$)

Which of the following is the correct expression for the electrode potential of a cell?

For the cell reaction $Zn_{(s)} + 2H^+_{(aq)} \to Zn^{2+}_{(aq)} + H_{2(g)}$,what happens when $H_2SO_4$ is added to the cathode compartment?

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