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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Explain the equilibrium state in a Daniell cell and derive its equilibrium constant. Given: $E^o_{cell} = 1.1 \ V$.

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Consider the electrochemical cell: $Pt \ | \ O_{2(g)} \ (1 \ bar) \ | \ HCl \ (aq) \ || \ M^{2+} \ (aq, 1.0 \ M) \ | \ M_{(s)}$. The pH above which, oxygen gas would start to evolve at the anode is . . . . . . (nearest integer). $\left[ \text{Given :} \ E^{\circ}_{M^{2+}/M} = 0.994 \ V, \ E^{\circ}_{O_{2}/H_{2}O} = 1.23 \ V, \ \frac{RT}{F}(2.303) = 0.059 \ V \ \text{at the given condition} \right]$

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