JEE Advanced 2022 · Paper 1 · Chemistry Q2

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ChemistryNumerical+3Question 38 of 54

The reduction potential ( $E^0$ , in V) of $\ce{MnO4^-(aq)}/\ce{Mn(s)}$ is _____.

[Given: $E^0_{(\ce{MnO4^-(aq)}/\ce{MnO2(s)})} = 1.68$ V; $E^0_{(\ce{MnO2(s)}/\ce{Mn^2+(aq)})} = 1.21$ V; $E^0_{(\ce{Mn^2+(aq)}/\ce{Mn(s)})} = -1.03$ V]

Your answer

Answer and worked solution

Official answer: 0.74

Electrode potentials do not add, but $\Delta G=-nFE$ does. Add the three steps:

Step	$n$	$E^0$ (V)	$nE^0$
$\ce{MnO4- -> MnO2}$	3	1.68	5.04
$\ce{MnO2 -> Mn^2+}$	2	1.21	2.42
$\ce{Mn^2+ -> Mn}$	2	−1.03	−2.06

$E^0=\frac{5.04+2.42-2.06}{7}=\frac{5.40}{7}=0.77\ \text{V}.$

Topics:Gibbs energy and EMF Galvanic cells and EMF

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