Cr2O7^2-(aq) + HNO2(aq) → Cr^3+(aq) + NO3^-(aq)
Two half equations.
Cr2O7^2- + + 14H^+ + 6e ==> 2Cr^3+ + 7H2O
HNO2 + H2O ==> NO3^- + 2e + 3H^+
Multiply equation 1 by 1 and equation by 3 (to make the electrons equal), then add the two half cells. There will be some items that appear on both sides (H^+ will) so add or subtract from both sides to make H^+ appear on just one side. Post your completed work if you want me to check it.
Balance the following redox reaction in acidic conditions
Cr2O7
2-
(aq) + HNO2 (aq) → Cr3+(aq) + NO3
-
(aq)
2 answers
Best
2 answers