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Nernst Equation

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Question

Calculate emf of a cell Ag(s) $/ {Ag}^{+}$ (1M) / ${Mg}^{+ 2}$ (1M) / Mg(s) and also predict the spontaneity of cell reaction. $(E_{{Ag}^{+} / Ag}^{0} = + 0 .8 V; E_{{Mg}^{+ 2} / Mg}^{o} = - 2 .37 V)$

Moderate

Solution

LHE ….Anode
RHE ….Cathode
Both the values given are SRP values
$∴ E_{cell}^{o} = E_{ca thode}^{o} - E_{anode}^{o}$
= (–2.37V)–(+0.8V)
= –3.17V
$∴ E_{cell}^{0}$ is negative, the cell reaction is non-spontaneous

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Similar Questions

The reaction which occur in the Galvanic cell is
$\begin{matrix} {MnO}_{4}^{-} + 8 H^{+} + 5 {Fe}^{2 +} ⟶ {Mn}^{2 +} + 5 {Fe}^{3 +} + 4 H_{2} O \\ E^{\circ} ({MnO}_{4}^{-}, {Mn}^{2 +}, H^{+} ∣ Pt) = 1 .51 V and \\ E^{\circ} ({Fe}^{3 +}, {Fe}^{2 +} ∣ Pt) = 0 .77 V \end{matrix}$
(Q) How would the emf of the cell be increased above the standard emf ?

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