If $E^{\circ}(Zn_{(aq)}^{+2} \mid Zn_{(s)}) = -0.76 \ V$,calculate the potential for $Zn_{(s)} \rightarrow Zn_{(0.01 \ M)}^{+2} + 2e^{-}$ at $298 \ K$.

  • A
    $+0.8192 \ V$
  • B
    $-0.8192 \ V$
  • C
    $+0.7008 \ V$
  • D
    $-0.7008 \ V$

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Similar Questions

Which graph correctly correlates $E_{cell}$ as a function of concentrations for the cell (for different values of $M$ and $M'$):-
$Zn_{(s)} + Cu^{2+}(M) \to Zn^{2+}(M') + Cu_{(s)};$ $E^o_{cell} = 1.10 \, V$
$X$-axis : $log_{10} \frac{[Zn^{2+}]}{[Cu^{2+}]}$,$Y$-axis : $E_{cell}$

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The $emf$ (in $V$) of a $Daniell$ cell containing $0.1 \ M \ ZnSO_4$ and $0.01 \ M \ CuSO_4$ solutions at their respective electrodes is $\left(E_{Cu^{2+} / Cu}^{\circ}=+0.34 \ V ; E_{Zn^{2+} / Zn}^{\circ}=-0.76 \ V\right)$

For the cell,$Mn_{(s)} | Mn^{2+}_{(aq)} (0.4 \, M) || Sn^{2+}_{(aq)} (0.04 \, M) | Sn_{(s)}$,calculate the free energy change $(\Delta G)$ at $298 \, K$ in $kJ$. Given: $E^o_{Mn^{2+}/Mn} = -1.18 \, V$; $E^o_{Sn^{2+}/Sn} = -0.14 \, V$; $\frac{2.303RT}{F} = 0.06$.

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