Question

In: Chemistry

4. Given the following half reaction Fe(OH)3 (s) + e- +3H+ çè Fe2+ + 3H2O log...

4. Given the following half reaction

Fe(OH)3 (s) + e- +3H+ çè Fe2+ + 3H2O log K = 15.8

(6 pts) The Grinnell formation in Glacier N.P., formed from ocean sediments, is typified by contrasting gray and red colors, yet the total Fe (iron) concentration remains constant throughout the rock. Can you explain this color variation? (Hint: iron turns red as it “rusts”).

Solutions

Expert Solution

The given reaction is

Fe(OH)3 (s) + e- + 3 H+ ------> Fe2+ + 3 H2O log K = 15.8

Also, the total iron concentration remains constant. Now,

[Fe]total = [Fe2+] + [Fe3+]

Iron in rocks is usually oxidized to Fe3+ (since this is the more stable oxidation state of iron) and hence is easily converted to Fe(OH)3 on reaction with water. In the atmosphere, this Fe(OH)3 is easily converted to hydrated Fe2O3 which is rust. However, we are told that the total iron concentration remains constant. Therefore, we must have a significant conversion of Fe(OH)3 to Fe2+ as per the given reaction. Moreover, the equilibrium constant for the said reaction is highly positive (since log K has a high positive value) meaning that the equilibrium favours the right hand side. Therefore, as more iron in rocks is oxidized to Fe3+ and consequently converts to Fe(OH)3, the reduction of Fe(OH)3 to Fe2+ takes place as shown above. Fe2+ cannot stay bare in the atmosphere and quickly combines with O2 of air to form ferrous oxide (FeO), which is greyish-black in colour. This accounts for the grey and red colour of the rocks.

Since the total concentration of iron must stay constant, the Fe2+/Fe3+ oxides/hydroxides must stay in equilibrium and the two oxidation states of iron are easily interconvertible by loss or gain of an electron from the atmosphere.

The combination reaction is

2 Fe2+ + O2 + 2 e- -------> 2 FeO


Related Solutions

Balance redox reaction in basic solution: Fe(OH)2(s) + MnO4-(aq) -> MnO2 (s) + Fe(OH)3
Balance redox reaction in basic solution: Fe(OH)2(s) + MnO4-(aq) -> MnO2 (s) + Fe(OH)3
Derive a balanced equation for the reaction occurring in the cell: Fe(s)|Fe2+(aq)||Fe3+(aq),Fe2+(aq)|Pt(s) a.) If E?cell =...
Derive a balanced equation for the reaction occurring in the cell: Fe(s)|Fe2+(aq)||Fe3+(aq),Fe2+(aq)|Pt(s) a.) If E?cell = 1.21 V, calculate ?G? for the reaction. b.) If E?cell=1.21V, calculate the equilibrium constant for the reaction. c.) Use the Nernst equation to determine the potential for the cell: Fe(s)|Fe2+(aq,1.0�10-3M)||Fe3+(aq,1.0�10-3M),Fe2+(aq,0.10M)|Pt(s)
Consider the following half reactions at 298 K Fe2+ + 2 e- → Fe    Eo = -0.441...
Consider the following half reactions at 298 K Fe2+ + 2 e- → Fe    Eo = -0.441 V Cd2+ + 2 e- → Cd    Eo = -0.403 V A galvanice cell based on these half reactions is set up under standard conditions where each solution is 1.00 L and each electrode weighs exactly 100.0 g. How much will the Cd electrode weigh when the nonstandard potential of the cell is 0.02804 V?
Consider the following half reactions at 298 K Fe2+ + 2 e- → Fe    Eo = -0.441...
Consider the following half reactions at 298 K Fe2+ + 2 e- → Fe    Eo = -0.441 V Cd2+ + 2 e- → Cd    Eo = -0.403 V A galvanice cell based on these half reactions is set up under standard conditions where each solution is 1.00 L and each electrode weighs exactly 100.0 g. How much will the Cd electrode weigh when the nonstandard potential of the cell is 0.02880 V? The answer is 139g.
show the steps to balance the following equations Fe2(SO4)3 + KOH = K2SO4 + Fe(OH)3
show the steps to balance the following equations Fe2(SO4)3 + KOH = K2SO4 + Fe(OH)3 and C6H10 + O2 = CO2 + H2O  
FeCl3(aq) + 3 NaOH (aq) ---- Fe(OH)3(s) + 3 NaCl(aq) Is this a redox reaction in...
FeCl3(aq) + 3 NaOH (aq) ---- Fe(OH)3(s) + 3 NaCl(aq) Is this a redox reaction in the reaction above? Explain.
For the following reaction: 3H2SO4(g) + 2Fe(OH)3(s) → Fe2(SO4)3(s) + 6H2O(l) 1. Calculate ΔH°rx (in kJ)...
For the following reaction: 3H2SO4(g) + 2Fe(OH)3(s) → Fe2(SO4)3(s) + 6H2O(l) 1. Calculate ΔH°rx (in kJ) for this reaction. Report your answer to two decimal places in standard notation (i.e. 123.45 kJ). 2. Calculate ΔS°rx (in J/K) for this reaction. Report your answer to two decimal places in standard notation (i.e. 123.45 J/K). 3. Calculate ΔG°rx (in kJ) at 511.71 K for this reaction. Report your answer to two decimal places in standard notation (i.e. 123.45 kJ). Assume ΔH°f and...
Consider an electrochemical cell based on the following overall reaction, Fe(s) + 2Ag+(aq)  Fe2+(aq) +...
Consider an electrochemical cell based on the following overall reaction, Fe(s) + 2Ag+(aq)  Fe2+(aq) + 2Ag(s) Fe2+(aq) + 2e– Fe(s), ℰ° = –0.44 V Ag+(aq) + e– Ag, ℰ° = 0.80 V Calculate the cell potential (in V) for this reaction at 25oC when the concentration of Ag+ ions is 0.050 M and the concentration of Fe2+ ions is 1.50 M. a. +1.32 V b. +1.50 V c. +1.20 V d. -1.32 V e. -1.20 V
Balance each of the following reations using the half reaction method. Show all work a. Bi(OH)3(S)...
Balance each of the following reations using the half reaction method. Show all work a. Bi(OH)3(S) + Sn(OH)3-(aq) ---> Sn(OH)62- (aq) + Bi(s) (Basic solution) b. BrO3-(aq) + Sb3+(aq) ---> Br - (aq) + Sb5+ (aq) (acidic solution) Thank you for your help!
Consider the reaction Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s) at 89 ∘C , where [Fe2+]= 3.00 M and [Mg2+]= 0.110 M...
Consider the reaction Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s) at 89 ∘C , where [Fe2+]= 3.00 M and [Mg2+]= 0.110 M . Part A: What is the value for the reaction quotient, Q, for the cell? Part B: What is the value for the temperature, T, in kelvins? Part C: What is the value for n? Part D: Calculate the standard cell potential for Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s)
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT