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Example · Example 6

Q.Using the standard reduction potentials E∘(Cu2+/Cu)=+0.34 VE^{\circ}(\text{Cu}^{2+}/\text{Cu}) = +0.34\ \text{V} and E∘(Zn2+/Zn)=−0.76 VE^{\circ}(\text{Zn}^{2+}/\text{Zn}) = -0.76\ \text{V}, calculate the standard EMF, Ecell∘E^{\circ}_{cell}, of the Daniell cell.

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The standard EMF of a galvanic cell is calculated as Ecell∘=Ecathode∘−Eanode∘E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}, where both potentials are taken as standard reduction potentials (never reversed for the anode). Copper has the more positive reduction potential, so it is the cathode; zinc has the more negative reduction potential, so it is the anode. Substituting, Ecell∘=E∘(Cu2+/Cu)−E∘(Zn2+/Zn)=0.34 V−(−0.76 V)=0.34+0.76=1.10 VE^{\circ}_{cell} = E^{\circ}(\text{Cu}^{2+}/\text{Cu}) - E^{\circ}(\text{Zn}^{2+}/\text{Zn}) = 0.34\ \text{V} - (-0.76\ \text{V}) = 0.34 + 0.76 = 1.10\ \text{V}. This positive value confirms the Daniell cell reaction is spontaneous as written (zinc displacing copper), consistent with the everyday observation that a zinc rod dipped in copper sulfate solution becomes coated with metallic copper. [!ANSWER] Ecell∘=1.10 VE^{\circ}_{cell} = 1.10\ \text{V}.

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