Which of the following expressions is used to | vedclass

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(A) The cell reaction is as follows:$Cd_{(s)} \longrightarrow Cd_{(aq)}^{2+} + 2e^{-}$(Oxidation at anode)$Cu_{(aq)}^{2+} + 2e^{-} \longrightarrow Cu_

Vedclass · Accepted Answer

$E_{cell} = E_{cell}^{o} - 0.0296 \log \frac{[Cd^{2+}]}{[Cu^{2+}]}$

Vedclass · Answer

(A) The cell reaction is as follows:$Cd_{(s)} \longrightarrow Cd_{(aq)}^{2+} + 2e^{-}$(Oxidation at anode)$Cu_{(aq)}^{2+} + 2e^{-} \longrightarrow Cu_{(s)}$(Reduction at cathode)$Cd_{(s)} + Cu_{(aq)}^{2+} \longrightarrow Cd_{(aq)}^{2+} + Cu_{(s)}$(Overall cell reaction)Using the Nernst equation at $25^{\circ} C$ $(298 \ K)$:$E_{cell} = E_{cell}^{o} - \frac{0.0592}{n} \log \frac{[Product]}{[Reactant]}$Here,the number of electrons transferred $n = 2$.Substituting the concentrations:$E_{cell} = E_{cell}^{o} - \frac{0.0592}{2} \log \frac{[Cd^{2+}]}{[Cu^{2+}]}$$E_{cell} = E_{cell}^{o} - 0.0296 \log \frac{[Cd^{2+}]}{[Cu^{2+}]}$

Which of the following expressions is used to find the cell potential of the $Cd_{(s)} | Cd_{(aq)}^{2+} || Cu_{(aq)}^{2+} | Cu_{(s)}$ cell at $25^{\circ} C$?

Similar Questions

The 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 :
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For the cell $Zn | Zn^{2+}(0.01 \, M) || Fe^{2+}(0.001 \, M) | Fe$ at $25^o C$,the $E_{cell} = 0.2905 \, V$. The equilibrium constant $K_c$ is:

Write the Nernst equation for the $E_{cell}$ of a Daniell 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 $\left(E_{Cu^{2+} / Cu}^{\circ}=+0.34 \ V ; E_{Zn^{2+} / Zn}^{\circ}=-0.76 \ V\right)$

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