Write the Nernst equation and calculate the emf of the following cell at 298 K : Zn|Zn2+ (0·001 M)||H+ (0·01 M)|H2 (g) (1 bar)|Pt (s) Given : E⊖Zn2+/Zn = – 0·76 V, E⊖H+/H2 = 0·00 V [log 10 = 1]

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TextNernst Equation for Electrode and Cell PotentialCBSE PYQ

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Calculate the e.m.f. of the following cell at 298K : Fe(s) | Fe²⁺ (0.001 M) || H⁺ (0.01M) | H₂(g) (1 bar) | Pt(s) Given that E°cell = +0.44 V [log 2 = 0.3010 log 3 = 0.4771 log 10 = 1]

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Nernst Equation for Electrode and Cell Potential

Variation of Conductivity with Concentration

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