Maharashtra BoardEnglish MediumSTD 12 ScienceChemistryElectrochemistry2 Marks
Question
State (or write) Nernst equation for the electrode potential and explain the terms involved.
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Answer
The Nernst equation for the single electrode reduction potential for a given ionic concentration in the solution in the case, $M_{(a q)}^{n+}+ ne ^{-} \rightarrow M _{( s )}$ is given by $\begin{aligned} & E_{ M ^{n+} / M }=E_{ M ^{ n } / M }^0-\frac{2.303 R T}{n F} \log _{10} \frac{[ M ]}{\left[ M ^{ n +}\right]} OR \\ & E_{ M ^{ n } / M }=E_{ M ^{n^{+} / M }}^0-\frac{2.303 R T}{n F} \log _{10} \frac{[1]}{\left[ M ^{ n +}\right]} \end{aligned}$ $E _{ M ^{ n +} / M }$ is the single electrode potential, $E_{ M ^{n+} / M }^0$ is the standard reduction electrode potential, $R$ is the gas constant $=8.314 JK ^{-1} mol ^{-1}$ $T$ is the absolute temperature, $n$ is the number of electrons involved in the reaction, F is Faraday ( 96500 C) $\left[ Mn ^{+}\right]$is the molar concentration of ions.
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