If a solution of $Cu^{+2}/Cu$ at $298 \, K$ is diluted $100$ times,how will the electrode potential change?

  • A
    $59 \, mV$ increase
  • B
    $59 \, mV$ decrease
  • C
    $29.5 \, mV$ increase
  • D
    $29.5 \, mV$ decrease

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Similar Questions

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). $\left[ \text{Given :} \ E^{\circ}_{M^{2+}/M} = 0.994 \ V, \ E^{\circ}_{O_{2}/H_{2}O} = 1.23 \ V, \ \frac{RT}{F}(2.303) = 0.059 \ V \ \text{at the given condition} \right]$

Calculate the potential of a hydrogen electrode in contact with a solution whose $pH = 10$.

What pressure $(bar)$ of $H_2$ would be required to make the $emf$ of a hydrogen electrode zero in pure water at $25^{\circ} C$?

The Gibbs energy change of the reaction (in $kJ \ mol^{-1}$) corresponding to the following cell $Cr | Cr^{3+} (0.1 \ M) || Fe^{2+} (0.01 \ M) | Fe$ is: (Given: $E^{\circ}_{Cr^{3+}/Cr} = -0.74 \ V$,$E^{\circ}_{Fe^{2+}/Fe} = -0.44 \ V$)

For the cell at $298 \ K$:
$Ag_{(s)} | AgBr_{(s)} | Br^{-}(0.01 \ M) || I^{-}(0.02 \ M) | AgI_{(s)} | Ag_{(s)}$
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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