For a cell involving a one-electron change at $25^o C$,$E^{o}_{cell} = 0.591 \ V$. The equilibrium constant for the reaction is .....

  • A
    $1 \times 10^{10}$
  • B
    $1 \times 10^{5}$
  • C
    $1 \times 10^{1}$
  • D
    $1 \times 10^{30}$

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At $298 \ K$,find out the $emf$ for the cell:
$Al_{(s)} | Al^{+3} (0.1 \ M) || Fe^{+2} (0.001 \ M) | Fe_{(s)}$
Given: $E^o_{Al^{+3}/Al} = -1.66 \ V$ and $E^o_{Fe^{+2}/Fe} = -0.44 \ V$.

Calculate the equilibrium constant of the reaction:
$Cu_{(s)} + 2Ag^{+}_{(aq)} \rightarrow Cu^{2+}_{(aq)} + 2Ag_{(s)}$
Given $E^{\Theta}_{cell} = 0.46 \ V$

The equilibrium constant for the following general reaction is $10^{30}$. Calculate $E^o$ for the cell at $298 \ K$.
$2 X_2(s)+3 Y^{2+}(a q) \rightarrow 2 X_2^{3+}(a q)+3 Y(s)$

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