In the reaction Cu+ + Fe---> Cu + Fe2+?
828740
fe is the reducing agent cos fe undergo oxidation prosess changing its oxidation no from 0 to 2+
825401
First of all I assume  Na2Co3 should be Na2CO3.\n\nFe(CO)5 + 4NaOH = Na2Fe(CO)4 + Na2CO3+ 2H2O\n\nFirst, balance out the metals (Na and Fe) then C, which was already balanced. Then I chose to balance out H and which automatically balanced out O.
343278
Half reactions are:\nOxidation: Cu ----> Cu2+ + 2e    Eo = -0.34V\nReduction: NO3- + 4H+ + 3e --->  NO + 2H2O   Eo = +0.96V\n\nAdd the two voltages (do not multiply this number by the coefficients you use to cancel the number of electrons out) and you get the +0.62V for the Eo of the cell.\n\nThe Nernst equation will take this form:\nE = Eo - 0.0592/n * log {(PNO)^2 * [Cu2+]^3 / [NO3-]^2 * [H+]^8}\n\nn will equal 6 here, not 8, as that will be the number of electrons that will be cancelled out when you add the oxidation and reduction reactions together (multiplied by the correct number to cancel those electrons out).\n\nThe concentration's portion simplifies, once you plug all of the numbers in, to about 18.94, whose log is 1.278.  Multiply that by 0.0592 and divide by 6 and you get 0.01, so only 0.01 V will be lost from the standard potential for that reaction at those concentrations.
770385
FeCl3+3NaOH>>Fe(OH)3+3NaCl
876667
Fe + CuCl2 --> FeCl2 + Cu\n\nmole of CuCl2 = 8.00 / 134.5 = 59.47 mmole\n\nmole of Fe = 5.40 / 56 = 96.42 mmole\nmole of unreacted Fe = 4.08 / 56 = 72.85 mmole\nmole of reacted Fe = 96.42 - 72.85 = 23.56 mmole\n\nmole of unreacted CuCl2 = 59.47 - 23.56 = 35.09 mmole\n\nYep, there will be around 35 mmole CuCl2 left in the beaker.\n\npeace\nvixklen
893153
2Fe (s) + 3F2 (g) ---> 2FeF3(s)
918374
Fe2O3 -- Fe+O2\n\nthis is the balanced form\n\n2(Fe2O3)--- 4Fe+3O2\nwithout the parentheses
831216
Cu (s) + Pb(NO3)2 (aq) ---------> No Reaction. \nCopper will not replace lead.\n\nreaction VI in the attached link.
888667
Cu2S + 12HNO3 -> Cu(NO3)2 + CuSO4 + 10NO2 + 6H2O\n\nThere you go.
374219
For the cell reaction: FeO(s) + H2(g) = Fe(s) + H2O(l) \n\nDG0(298 K) = DGf0 Fe(s) + DGf0 H2O(l) – DGf0 FeO(s) – DGf0 H2(g)\n\nFrom any table of Thermodynamic Properties:\nDGf0 Fe(s) = 0 KJ/mol\nDGf0 H2O(l) = -237.192 KJ/mol\nDGf0 FeO(s) = -251.4 KJ/mol\nDGf0 H2(g) = 0 KJ/mol\n\nThen: \nDG0(298 K) = 14.204 KJ/mol\n\nFurthermore: E0 = -DG0/(nF) = -14204/(2 x 96500) = -0.074 V vs NHE (normal hydrogen electrode), where n: number of electrons in each half reaction (FeO/Fe and 2H+/H2, n=2) and F is the Faraday constant (96500 C/mol).
861343
Ferrous means fe (II)\n1) 3Fe2+ + MnO4- + 4H+ to give  3Fe3+ + Mn2+ + 2H2O \n\nThere are two half equations\nthe first is pretty simple: 3 Fe2+ to give 3 Fe3+. That is the oxidation reaction\n\nthe other is the reduction. \nMn +7 is reduced to Mn+2\nthe secret to balancing this is to add water to the side with no oxygen, and H+ to the other side to balance it out. Check that the charges are balanced\n\n2) Fe3+ + Sn2+ to give Fe2+ + Sn3+\n \n3) cant help u here u need a textbook!
612107
a. 2k+ = .20moles , SO42-=.10moles\n\nb.Fe3+=.10moles,3NO3-= .30moles\n\nc.2Al3+=.20moles,3SO42-=.30moles\n\nd. Ni2+=.10moles,SO42-=.10moles
929071
2C18H18 + 45 O2 --->  36 CO2 + 18 H2O\n3Cu + 8HNO3 ----> 3Cu(NO3)2 + 2NO + 4H2O
372026
CuSO4·5H2O(s) → CuSO4(s) + 5H2O(g)
801118
CuCl2 + 2NH4OH --> Cu(OH)2 + 2NH4Cl\n\nCu(OH)2 (s) + 4NH3 (aq)     ----->     Cu(NH3)4^2+
411418
Ox reaction:  Fe ---> Fe(2+) + 2e\nRed reaction:  Ni(2+) + 2e ---> Ni\n\nGiven that Fe/Fe(2+) is +0.44V and Ni(2+)/Ni is -0.26V, the net voltage is +0.18V.
367815
Try finding the standard reduction potential for Cu and NO first then use those poteintial rxn equ to form the above full reaction afterwhich simply add or substract the poteinital u find to get E for the above reaction. \n\nSince E is proportional to the rate constate, u can use the above concentration as well as partial pressure to calculate the rate constant and hence E. Taking care that rate constant is the concentration/partial pressure of the reactant over the product!