(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.
Explanation
(a) From +6 to +3
(b) The reaction involves the transfer of 6 electrons from 6Fe 2+ to Cr\(_2\)O\(_7\)
SO, 6Fe\(^{2+}\) \(\rightleftharpoons\) 6Fe\(^{3+}\) + 6e\(^-\)
The addition of the 2 half cell equations gives Cr\(_2\)O\(_7^{2-}\) + 14H\(^+\) — 6Fe\(^{2+}\) \(\to\) 2Cr\(^{3+}\) + 6Fe\(^{3+}\) + 6Fe\(^{3+}\) + 7H\(_2\)O.