(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.