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 $25^{\circ} C$?

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