Nernst Equation

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Question

Emf of the cell $Zn / {Zn}^{+ 2} / / {Cu}^{+ 2} / Cu$ will be highest when

Moderate

Solution

$E_{cell} = E_{cell}^{o} - \frac{0 .0591}{n} \log \frac{[Products]}{[Reactants]}$
$Anode - {Zn}^{} \to {Zn}^{+ 2} + 2 e^{-}$
$Cathode {Cu}^{+ 2} + 2 e^{-} \to Cu$
$Zn + {Cu}^{+ 2} \to {Zn}^{+ 2} + Cu$
${E^{0}}_{cell} is a constant$
$E_{cell} \propto - \log \frac{[{Zn}^{+ 2}]}{[{Cu}^{+ 2}]}$
$Lesser the \frac{[{Zn}^{+ 2}]}{[{Cu}^{+ 2}]}, higher will be the E_{cell}$
$∴ (2) is correct$

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Nernst Equation

Unlock the full solution & master the concept.

Emf of the cell Zn /Zn+2//Cu+2/Cu will be highest when

Ecell = Ecello−0.0591nlog[Products]ReactantsAnode−Zn → Zn+2+2e−Cathode Cu+2 +2e-→Cu Zn + Cu+2 → Zn+2 +CuE0cell is a constantEcell∝ −logZn+2Cu+2Lesser theZn+2Cu+2,higher will be the Ecell∴ (2) is correct

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Emf of the cell $Zn / {Zn}^{+ 2} / / {Cu}^{+ 2} / Cu$ will be highest when