The zirconium coating on the uranium rods is present to prevent the uranium trom
ID: 524254 • Letter: T
Question
The zirconium coating on the uranium rods is present to prevent the uranium trom reacting wwah other materials. Below is an example of how a uranium compound can react with dichromate ion. -0.163V, a. Write a balanced reduction half reaction for eq 2 in acidic media (A pts) b. Write a balanced oxidation half reaction for eq 2 in acidic media (4 pts) c. Write the overall balanced redox reaction Include phase labels. (2 pts) d. Now give the balanced redox reaction as if the reaction were performed in basic media. Include phase labels. (2 pts) suppose the system is 0.5 M UO o.2 M uoz2. and o.2 M crzo o.1 M cra What would the electrochemical cell potential be for a spontaneous system? (E-E o.05916ln vogton a pt The zirconium coating on the uranium rods is present to prevent the uranium trom reacting wwah other materials. Below is an example of how a uranium compound can react with dichromate ion. -0.163V, a. Write a balanced reduction half reaction for eq 2 in acidic media (A pts) b. Write a balanced oxidation half reaction for eq 2 in acidic media (4 pts) c. Write the overall balanced redox reaction Include phase labels. (2 pts) d. Now give the balanced redox reaction as if the reaction were performed in basic media. Include phase labels. (2 pts) suppose the system is 0.5 M UO o.2 M uoz2. and o.2 M crzo o.1 M cra What would the electrochemical cell potential be for a spontaneous system? (E-E o.05916ln vogton a ptExplanation / Answer
a. Balanced equation for equation 2 in acidic medium
3UO^2+ + Cr2O7^2- + 8H+ ------> 3UO2^2+ + 2Cr3+ + 4H2O
b. balanced oxidation half reaction,
UO^2+ ---> UO2^2+ + 2e-
c. Overall balanced redox reaction with phases,
3UO^2+ (aq) + Cr2O7^2- (aq) + 8H+ (aq) ------> 3UO2^2+ (aq) + 2Cr3+ (aq) + 4H2O(l)
d. Balanced equation in basic medium,
3UO^2+ (aq) + Cr2O7^2- (aq) + 4H2O (l) ------> 3UO2^2+ (aq) + 2Cr3+ (aq) + 8OH- (aq)
e. Using Nernst equation,
Ecell = Eo - 0.0592/n logK
= (1.232 - 0.062) - 0.0592/6 log[(0.1)^2.(0.2)^3/(1)^8.(0.2).(0.5)^3]
= 1.195 V
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