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Electrochemical Series

When two half...

Question

When two half cells of electrode potential $E_{1}$ and $E_{2}$ combine to form a cell of electrode potential $E_{3}$ then choose the correct relation :
Here
$n_{1}, n_{2} & n_{3} are no. of electrons exhanged in first, second and combined half cells$

E_{3} = E_{1} - E_{2}

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E_{3} = E_{1} + E_{2}

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E_{3} = \frac{n_{1} E_{1} + n_{2} E_{2}}{n_{3}}

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E_{3} = \frac{E_{1} + E_{2}}{n_{1} n_{2} n_{3}}

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Solution

The correct option is C $E_{3} = \frac{n_{1} E_{1} + n_{2} E_{2}}{n_{3}}$
Taking example :
$F e^{3 +} (a q) + e^{-} ⇌ F e^{2 +} (a q)$
$Electrode potential = E_{1}$

$F e^{2 +} (a q) + 2 e^{-} ⇌ F e (s)$
$Electrode potential = E_{2}$

Combined:
$F e^{3 +} (a q) + 3 e^{-} ⇌ F e (s)$
$Electrode potential = E_{3}$

Since
$Δ G = - n F E_{c e l l}$
$∴ - n_{1} F E_{1} - n_{2} F E_{2} = - n_{3} F E_{3} E_{3} = \frac{n_{1} E_{1} + n_{2} E_{2}}{n_{3}}$

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