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Exercise · Q8

Q.Using E∘(Ag+/Ag)=+0.80 VE^{\circ}(\text{Ag}^{+}/\text{Ag}) = +0.80\ \text{V} and E∘(Cu2+/Cu)=+0.34 VE^{\circ}(\text{Cu}^{2+}/\text{Cu}) = +0.34\ \text{V}, predict whether metallic copper placed in a solution of AgNO3\text{AgNO}_3 will spontaneously reduce Ag+\text{Ag}^+ ions. Justify using the sign of Ecell∘E^{\circ}_{cell}.

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For copper metal to reduce silver ions, the relevant reaction is Cu(s)+2Ag+(aq)→Cu2+(aq)+2Ag(s)\text{Cu}(s) + 2\text{Ag}^{+}(aq) \to \text{Cu}^{2+}(aq) + 2\text{Ag}(s), in which Ag+\text{Ag}^+ is reduced (cathode process) and Cu\text{Cu} is oxidized (anode process). Its standard EMF is Ecell∘=Ecathode∘−Eanode∘=E∘(Ag+/Ag)−E∘(Cu2+/Cu)=0.80 V−0.34 V=+0.46 VE^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = E^{\circ}(\text{Ag}^{+}/\text{Ag}) - E^{\circ}(\text{Cu}^{2+}/\text{Cu}) = 0.80\ \text{V} - 0.34\ \text{V} = +0.46\ \text{V}. Because Ecell∘E^{\circ}_{cell} is positive, ΔG∘=−nFEcell∘\Delta G^{\circ} = -nFE^{\circ}_{cell} is negative, and the reaction is thermodynamically spontaneous as written. This is the chemistry behind the classic 'silver tree' demonstration, in which a co …

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