Write the Nernst equation for the $E_{cell}$ of a Daniell cell.
(N/A) The Daniell cell reaction is given by: $Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$.
For this cell,the Nernst equation at $298 \ K$ is expressed as:
$E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$.
Here,$n = 2$ (number of electrons transferred in the redox reaction).
Thus,the equation becomes: $E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{2} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$.
For this cell,the Nernst equation at $298 \ K$ is expressed as:
$E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$.
Here,$n = 2$ (number of electrons transferred in the redox reaction).
Thus,the equation becomes: $E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{2} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$.
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What is the potential of the cell containing two hydrogen electrodes as represented below $V$:
$Pt | \frac{1}{2} H_{2(g)} | H^{+} (10^{-8} M) || H^{+} (10^{-3} M) | \frac{1}{2} H_{2(g)} | Pt$
$Pt | \frac{1}{2} H_{2(g)} | H^{+} (10^{-8} M) || H^{+} (10^{-3} M) | \frac{1}{2} H_{2(g)} | Pt$
Medium
View SolutionWhat is $E_{cell}$ (in $V$) of the following cell at $298 \ K$ ?
$(E^{\ominus}_{Zn^{2+}/Zn} = -0.76 \ V ; E^{\ominus}_{Ni^{2+}/Ni} = -0.25 \ V ; \frac{2.303 RT}{F} = 0.06 \ V)$
$Zn_{(s)} | Zn^{2+} (0.01 \ M) || Ni^{2+} (0.1 \ M) | Ni_{(s)}$
$(E^{\ominus}_{Zn^{2+}/Zn} = -0.76 \ V ; E^{\ominus}_{Ni^{2+}/Ni} = -0.25 \ V ; \frac{2.303 RT}{F} = 0.06 \ V)$
$Zn_{(s)} | Zn^{2+} (0.01 \ M) || Ni^{2+} (0.1 \ M) | Ni_{(s)}$
$Cu^{2+} + 2e^- \to Cu$. On increasing $[Cu^{2+}]$ concentration,electrode potential
Medium
View SolutionWhich of the following relations represents the correct relation between standard electrode potential and equilibrium constant?
$I$. $\log K = \frac{nF E^o}{2.303 RT}$
$II$. $K = e^{\frac{nF E^o}{RT}}$
$III$. $\log K = -\frac{nF E^o}{2.303 RT}$
$IV$. $\log K = 0.4342 \frac{nF E^o}{RT}$
Choose the correct statement$(s)$.
$I$. $\log K = \frac{nF E^o}{2.303 RT}$
$II$. $K = e^{\frac{nF E^o}{RT}}$
$III$. $\log K = -\frac{nF E^o}{2.303 RT}$
$IV$. $\log K = 0.4342 \frac{nF E^o}{RT}$
Choose the correct statement$(s)$.
MediumAIIMS 2017
View SolutionThe standard $e.m.f.$ of a cell,involving one electron change is found to be $0.591 \ V$ at $25^{\circ} C$. The equilibrium constant of the reaction is :
$(F=96500 \ C \ mol^{-1} ; R=8.314 \ JK^{-1} \ mol^{-1})$
$(F=96500 \ C \ mol^{-1} ; R=8.314 \ JK^{-1} \ mol^{-1})$
Medium
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