Calculate $E_{cell}$ for the following cell:
$Pt_{(s)} | H_{2(g)} | HA, 1 \ M \ (K_a = 10^{-7}) || HB, 1 \ M \ (K_a = 10^{-5}) | H_{2(g)} | Pt_{(s)}$ (in $V$)

  • A
    $0.06$
  • B
    $0.03$
  • C
    $0.04$
  • D
    $0.05$

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

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The correct information is:
[Given: $K_{sp}(AgBr) = 4 \times 10^{-13}$,$K_{sp}(AgI) = 8 \times 10^{-17}$,$\frac{2.303 \ RT}{F} = 0.06 \ V$,$\log 2 = 0.3$]

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What is the change in the reduction potential of a hydrogen electrode when the $pH$ of the initial solution changes from $0$ to $7$ by neutralization?

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has $E^{\Theta}_{cell} = 0.236 \ V$ at $298 \ K$. Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.

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