The initial $EMF$ of the concentration cell shown in the figure at $298 \ K$ is .............. $V$.

  • A
    $-0.059$
  • B
    $0.059$
  • C
    $0.59$
  • D
    $0.0059$

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Similar Questions

What will be the electromotive force of the following cell at $298 \ K$ ............. $V$
$Pt \, | \, Br_{2(\ell), 0.1M} \, | \, Br^{-}_{(aq), 0.1M} \, || \, H^{+}_{(aq), 0.1M} \, | \, H_{2(g), 1atm} \, | \, Pt$
Given: $E^{0}_{Br^{-}/Br_{2}} = -1.06 \ V$

For an electrochemical cell
$Sn_{(s)} | Sn^{2+}(aq, 1 \ M) || Pb^{2+}(aq, 1 \ M) | Pb_{(s)}$
the ratio $\frac{[Sn^{2+}]}{[Pb^{2+}]}$ when this cell attains equilibrium is
(Given $E^{0}_{Sn^{2+}/Sn} = -0.14 \ V$,$E^{0}_{Pb^{2+}/Pb} = -0.13 \ V$,$\frac{2.303 \ RT}{F} = 0.06$)

$A$ $Daniel$ cell is made at $25\,^oC$ by connecting $Zn/Zn^{2+} (0.1\ M, 1\ L)$ and $Cu/Cu^{2+} (0.9\ M, 1\ L)$ electrodes. The cell is discharged until its $emf$ reaches $1.10\ V$,then it is charged (reversing the discharge process) by passing $0.6\ F$ charge.
[Given: ${E^0}_{Zn^{2+}/Zn} = -0.76\ V, {E^0}_{Cu^{2+}/Cu} = 0.34\ V, \frac{2.303RT}{F} = 0.06, \log 2 = 0.30]$
Select the incorrect option.

For a Daniell cell,$E^0_{cell} = 1.1 \ V$. How is $K_c$ represented for the reaction occurring in the Daniell cell?

The standard electrode potential of a $Cu^{2+} | Cu$ electrode is $0.34 \, V$ (reduction potential). What will be the electrode potential of a $0.001 \, M \, Cu^{2+}$ solution in $V$?

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