The e.m.f. of the cell in which the following reaction $Zn_{(s)} + Ni^{2+}(a = 1.0) \rightleftharpoons Zn^{2+}(a = 10) + Ni_{(s)}$ occurs,is found to be $0.5105 \ V$ at $298 \ K$. The standard e.m.f. of the cell is ............ $V$.

  • A
    $0.4810$
  • B
    $0.54$
  • C
    $0.5696$
  • D
    $-0.5105$

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Similar Questions

Calculate the electrode potential at $298 \; K$ for $Zn|Zn^{2+}$ electrode in which the activity of zinc ions is $0.001 \; M$ and $E^o_{Zn^{2+}/Zn}$ is $-0.76 \; V$. (in $; V$)

Calculate $E_{cell}^{\circ}$ if the equilibrium constant for the following reaction is $1.2 \times 10^6$.
$2 Cu_{(aq)}^{+} \longrightarrow Cu_{(aq)}^{2+} + Cu_{(s)}$ (in $V$)

What is the sign of $\Delta G^o$ and the value of $K$ for an electrochemical cell for which $E_{cell}^o = 0.80 \ V$?

If the standard electrode potential for a cell is $2 \ V$ at $300 \ K,$ the equilibrium constant $(K)$ for the reaction $Zn_{(s)} + Cu^{2+}_{(aq)} \rightleftharpoons Zn^{2+}_{(aq)} + Cu_{(s)}$ at $300 \ K$ is approximately $(R = 8 \ J \ K^{-1} \ mol^{-1}, F = 96000 \ C \ mol^{-1})$

For the galvanic cell,
$Zn_{(s)} + Cu^{2+}(0.02 \ M) \rightarrow Zn^{2+}(0.04 \ M) + Cu_{(s)}$
$E_{cell} = ...... \times 10^{-2} \ V \text{ (Nearest integer) }$
$[\text{Use}: E_{Cu^{2+}/Cu}^{0} = 0.34 \ V, E_{Zn^{2+}/Zn}^{0} = -0.76 \ V]$
$[\frac{2.303 \ RT}{F} = 0.059 \ V]$

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