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3 years ago

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Which process is an oxidation?

1 answer:

Zarrin [17]3 years ago

4 0

<span>The problem has to do with oxidation states of the matter. The oxidation state of oxygen will always be -2 with the exception of peroxides which will have a state of -1. The overall balanced state of chemical compounds will be 0, so the oxidation state of Mn in MnO2 will be +4. The oxidation state of MnO4- will then be +7 to balance out to the negative one charge. The state change from +4 to +7 is 3, thus three electrons have to be lost in order for this to happen; a loss of a charge of -3 results in an increase of charge of 3. Oxidation is always the process of 'losing' electrons.

</span><span>E] MnO2(s) MnO4-(aq</span>

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Explanation:

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$[FeSCN^{2+}] \rightleftharpoons [Fe^{3+}] + [SCN^{-}]$

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Now, first calculate the value of $K_{eq}$ as follows.

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Now, according to the concentration values at the re-established equilibrium the value for $[FeSCN^{2+}]$ will be calculated as follows.

$K_{eq} = \frac{[Fe^{3+}][SCN^{-}]}{[FeSCN^{2+}]}$

$11.24 \times 10^{-4} = \frac{8.12 \times 10^{-3} \times 7.84 \times 10^{-3}}{[FeSCN^{2+}]}$

$[FeSCN^{2+}] = 5.66 \times 10^{-2}$ M

Thus, we can conclude that the concentration of $[FeSCN^{2+}]$ in the new equilibrium mixture is $5.66 \times 10^{-2}$ M.

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