Chemistry related question

Question

Balance the equation for the reaction, using no fractional coefficients.

? Сr2O3+ ? КОН +? О2 -> ? K2CrO4 + ? H2O

The coefficient of KOH in the balanced equation is

(A) 2

(B) 4

(C) 6

(D) 8

(E)10

Answer 1

This is a redox reaction. You can tell immediately because oxygen (O2) goes from oxidation state of zero on the left to oxidation state of -2 on the right (in H2O and in K2CrO4). It has been reduced. What has been oxidized? Cr goes from 3+ in Cr2O3 to 6+ in K2CrO4. Cr has been oxidized. The reaction is taking place in basic conditions (indicated by presence of KOH).

2 Сr₂O₃ + 8 КОН + 3 О₂ -> 4 K₂CrO₄ + 4 H₂O .. balanced redox equation. See below for how to do this.

ANSWER = (D) 8

Half reactions:

O₂ + 4H₂O +4e- ==> 2H₂O + 4OH^- .. balanced reduction half reaction

Cr₂O₃ + 5H₂O + 10 OH^- ==> 2CrO₄^2- +10 H₂O + 6e- .. balanced oxidation half reaction

3O₂ + 12H₂O + 12e- ==> 6H₂O + 12OH^- ..

2Cr₂O₃ + 10H₂O + 20 OH^- ==> 4CrO₄^2- +20 H₂O + 12e-

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3O₂ + 2Cr₂O₃ + 8 OH^- ==> 4H₂O + 4CrO₄^2- .. balanced redox equation