swing the next reaction
Cr2O72-+I- ---> Cr3+I2
and write the coefficients
[ ] I-+ [ ]H++ [ ] Cr2O72- [ ] I2+[ ] H2O+[ ] Cr3+
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swing the next reaction Cr2O72-+I- ---> Cr3+I2 and write the coefficients [ ] I-+ [ ]H++...
Consider the reaction Cr2O72−(aq) + I−(aq) → Cr3+(aq) + I2(s).
In the reaction of dichromate ion and iodide ion: Cr2O72-(aq) + 14 H+(aq) + 6 I-(aq) ---> 2 Cr3+(aq) + 7 H2O(l)+ 3 I2 (s) the oxidizing agent is ? and the reducing agent is ?
In the redox reaction Cr2O72- (aq) + I1- (aq) ---> Cr3+ (aq) + I2 (s), what is the balanced oxidation reaction (before the half reactions are added together)?
Using the Nernst equation, calculate the cell potential for the following reaction (T=298 K): Cr2O72- (aq) + 14 H+ (aq) 6 I- (aq) → 2 Cr3+ (aq) + 3 I2 (s) + 7 H2O (l) given that Cr2O72- = 1.7 M H+ = 1 M I- = 1 M Cr3+ = 0.002 M
Balance the redox reaction by inserting the appropriate coefficients. redox reaction: H^+ + CrO2^−4 + NO−2⟶Cr3+ + H2O + NO^−3
Question 22 (4 points) Cr2O72- + 3 H2O2 + 8 H+ + 2 Cr3+ + 3 O2 + 7 H2O In the above reaction Ocrin Cr₂O7 2- is reduced. OH in H+ is oxidized. Cr₂O7 2 is the oxidizing agent and H2O2 is the reducing agent. Cr₂O7 2- is the reducing agent and H2O2 is the oxidizing agent.
Complete and balance the following half-reaction in basic solution Cr2O72-(aq) 2 Cr3+ (aq) 04- 03.02 + 2+ 3+ 4+ 1 3 4 5 N 6 7 8 9 0 12 n 4 Os 06 07 co 0, 0. + on (s) (1) (g) ((aq) O H2O ОН" H + CD H30+ С
Balance the redox reaction by inserting the appropriate coefficients. redox reaction: H+ + Cro- + NO3 + Cr3+ + H2O + NO3
Balance the following equations. (Use the lowest possible whole-number coefficients. These may be zero.) (a) MnO4-(aq) + Cl-(aq) Mn2+(aq) + Cl2(aq) MnO4- + Cl- + H+ + H2O Mn2+ + Cl2 + H+ + H2O (b) Cr2O72-(aq) + NO2-(aq) Cr3+(aq) + NO3-(aq) Cr2O72- + NO2- + H+ + H2O Cr3+ + NO3- + H+ + H2O (c) Tl2O3(s) + NH2OH(aq) TlOH(s) + N2(g) Tl2O3 + NH2OH + OH- + H2O TlOH(s) + N2 + OH- + H2O (d) CrO42-(aq) + C2O42-(aq) Cr(OH)3(s) + CO2(g) CrO42- + C2O42- + OH- + H2O Cr(OH)3 + CO2 + OH- + H2O
1) Balance the following reaction under acidic conditions and calculate the cell potential in (V) at 298 K generated by the cell. Report your answer to the hundredths place. Cr2O72-(aq) + I-(aq) → Cr3+(aq) + I2(s) [Cr2O72-] = 2.0 M, [H+] = 1.0 M, [I-] = 1.0 M, [Cr3+] = 1.0 × 10-5 M 2) What is the value of n for the following reaction? Enter the whole number. 3Ni+(aq) + Cr(OH)3(s) + 5OH-(aq) → 3Ni(s) + CrO42-(aq) + 4H2O(l)