(a) What is the change in oxidation state of chromium in the reaction represented by the following equation? 3SO\(_2\) + Cr\(_2\)O\(^2_{-7}\) + 2H\(^+\) -> ...
Question 1 Report
(a) What is the change in oxidation state of chromium in the reaction represented by the following equation?
3SO\(_2\) + Cr\(_2\)O\(^2_{-7}\) + 2H\(^+\) -> 3SO\(_4^{2-}\) + 2Cr\(^{3+}\) + H\(_2\)O
(b) Use the half equations given below to deduce the equation for the reaction between iron(II) ions and heptaoxodichromate (VI) ions in acidic solution.
Fe\(^{2+}\) --> Fe\(^{3+}\) + e\(^-\)
Cr\(_2\)O\(^{2-}_7\) + 14H\(^+\) + 6e\(^-\) ----> 2Cr\(^{3+}\) + 7H\(_2\)O.
Answer Details
(a) Change in oxidation state of chromium
In the heptaoxodichromate(VI) ion \(Cr_2O_7^{2-}\), each chromium has an oxidation number of +6 (oxygen is \(-2\): \(2x + 7(-2) = -2\Rightarrow x = +6\)).
In the product \(Cr^{3+}\), chromium is +3.
The oxidation state therefore decreases from +6 to +3, a change of 3 units. Chromium is reduced (each Cr gains 3 electrons).
(b) Deriving the overall equation
To cancel electrons, multiply the iron half-equation by 6 so that 6 electrons are supplied to match the 6 electrons gained by the dichromate:
\[6Fe^{2+} \to 6Fe^{3+} + 6e^-\]\[Cr_2O_7^{2-} + 14H^+ + 6e^- \to 2Cr^{3+} + 7H_2O\]Adding and cancelling the electrons gives the overall equation:
\[6Fe^{2+} + Cr_2O_7^{2-} + 14H^+ \to 6Fe^{3+} + 2Cr^{3+} + 7H_2O\]Download The App On Google Playstore
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