Chemistry - Electrochemistry Question with Solution | TestHub

ChemistryElectrochemistryGalvanic cells/Nernst Equation/Concentration CellsMedium2 minQB

ChemistryMediummultiple choice

In the following electrochemical cell:

Zn | Zn²⁺ || H⁺ | H₂(g) | Pt

Under what condition(s) does the $E_{ce ll}$ become equal to $E_{ce ll}^{\circ}$ ?

Answer:

A, B, C

Solution:

For cell:

Zn | Zn²⁺ || H⁺ | H₂

$E = E^{\circ} - \frac{0.0591}{2} lo g \frac{[ Z n ^{2 +} ] P _{H_{2}}}{[ H ^{+} ] ^{2}}$

When the reaction quotient $Q = 1$ , then $E_{ce ll} = E_{ce ll}^{\circ}$ .

Options (a), (b), (c) satisfy this condition.

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

⏱ 2mℹ️ Source: QB

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