What is the EMF of a galvanic cell if E cathode ∘ = 0.80 V and E anode ∘ = − 0.76 V?

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ChemistryElectrochemistryGalvanic cells/Nernst Equation/Concentration CellsEasy2 minPYQ_2026

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What is the EMF of a galvanic cell if $E_{cathode}^{\circ} = 0.80$ V and $E_{anode}^{\circ} = - 0.76$ V?

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Answer:

Solution:

The standard EMF of a galvanic cell is given by the difference between the standard reduction potentials of the cathode and the anode.

$E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}$

Substituting the given values:

$E_{cell}^{\circ} = 0.80 - (- 0.76)$

$E_{cell}^{\circ} = 0.80 + 0.76 = 1.56 V$

Answer: $1.56$ V.

Stream:MHT_CET_MEDICALSubject:ChemistryTopic:ElectrochemistrySubtopic:Galvanic cells/Nernst Equation/Concentration Cells

⏱ 2mℹ️ Source: PYQ_2026

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