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please answer in 10m 5. Balance each reaction, calculate the net Eº, and declare each reaction...
use the example to answer 8,9,10&11 Here's an example: Balance the following redox reaction, which occurs in acidic solution: Fe (aq)+ MnO4'(aq) -Fe (aq) + Mn (aq) Solution: +2 +7 +3 Step 1) +2 Fe2 (aq)+ MnOa (aq) Fe(aq) + Mn2 (aq) Fe (aq) MnO4 (aq) Mn2 (aq) 1 Fe on each side; 1 Mn on each side; no adjustment necessary Fe2 (aq) Fe 3'(aq) + e 5 e + MnO4(aq) Fe (aq) Fe (aq) + e (2+ on each...
Given the following standard reduction potentials Ru2+(aq) + 2e– → Ru(s) Eº = 0.46 V Pb2+(aq) + 2e– → Pb(s) Eº = –0.13 V Fe2+(aq) + 2e– → Fe(s) Eº = –0.44 V Cr3+(aq) + 3e– → Cr(s) Eº = –0.74 V Mn2+(aq) + 2e– → Mn(s) Eº = –1.19 V Mg2+(aq) + 2e– → Mg(s) Eº = –2.36 V choose all metals that will not prevent corrosion of iron by cathodic protection. Group of answer choices only Pb only...
use tabulated standard electrode potential to calculate the standard cell potential for the reaction occurring in an electrochemical cell at 25 C. (The equation is balanced.) 3Ni^2+(aq)+2Cr(s)--->3Ni(s)2Cr^3+(aq) Express your answer to two significant figures and include the appropriate units. em 26 E (V) -0.45 -0.50 -0.73 -0.76 -1.18 Standard reduction half-cell potentials at 25°C Half-reaction E° (V) Half-reaction Aul+ (aq) + 3e +Au(s) 1.50 Fe2+ (aq) + 2eFe(s) Ag+ (aq) +e-Ag(s) 0.80 Cr3+ (aq) + Cr²+ (aq) Fe+(aq) + 3e...
The equilibrium constant, K, for a redox reaction is related to the standard potential, Eº, by the equation In K = nFE° RT where n is the number of moles of electrons transferred, F (the Faraday constant) is equal to 96,500 C/(mol e), R (the gas constant) is equal to 8.314 J/(mol · K), and T is the Kelvin temperature. Standard reduction potentials Reduction half-reaction E° (V) Ag+ (aq) + e +Ag(s) 0.80 Cu²+ (aq) + 2e + Cu(s) 0.34...
A) Use tabulated electrode potentials to calculate ΔG∘ for the reaction. 2K(s)+2H2O(l)→H2(g)+2OH−(aq)+2K+(aq) B) (Refer to the following standard reduction half-cell potentials at 25∘C: VO2+(aq)+Ni2+(aq)2H+(aq)++2e−e−→ →Ni(s)VO2+(aq) +H2O(l)E∘=−0.23V E∘=0.99V) An electrochemical cell is based on these two half-reactions: Oxidation:Reduction:Ni(s)VO2+(aq,0.024M)+2H+(aq,1.4M)+e−→→Ni2+(aq,1.8M)+2e−VO2+(aq,1.8M)+H2O(l) Calculate the cell potential under these nonstandard concentrations. C) Standard reduction half-cell potentials at 25∘C Half-reaction E∘ (V ) Half-reaction E∘ (V ) Au3+(aq)+3e−→Au(s) 1.50 Fe2+(aq)+2e−→Fe(s) − 0.45 Ag+(aq)+e−→Ag(s) 0.80 Cr3+(aq)+e−→Cr2+(aq) − 0.50 Fe3+(aq)+3e−→Fe2+(aq) 0.77 Cr3+(aq)+3e−→Cr(s) − 0.73 Cu+(aq)+e−→Cu(s) 0.52 Zn2+(aq)+2e−→Zn(s) − 0.76...
Please combine the Mn electrode half reaction with the Ag electrode half reaction and write the complete redox reaction using date from table 8.1. Indicate which is the oxidizing and which is the reducing agent. b. Calculate the EMF when the reaction quotient (Q) = ]Mn2+]/[Ag+]^2 = 10^-5 Reducing agent Half-reaction E0 (volts) -2.93 2.87 Ca Na Mg Mn Zn Fe Ni Pb Ca Ca2++2e Na → Na++e- -2.36 Mn Mn++2e Zn Zn2++2e Fe → Fe2++ 2e- NiNi2++2e- -0.76 -0.47...
Use the tabulated electrode potentials to calculate K for the oxidation of nickel by H+: Ni(s)+2H+(aq)→Ni2+(aq)+H2(g) Express your answer using two significant figures. Standard reduction half-cell potentials at 25∘C Half-reaction E∘ (V) Half-reaction E∘ (V) Au3+(aq)+3e−→Au(s) 1.50 Fe2+(aq)+2e−→Fe(s) −0.45 Ag+(aq)+e−→Ag(s) 0.80 Cr3+(aq)+e−→Cr2+(aq) −0.50 Fe3+(aq)+3e−→Fe2+(aq) 0.77 Cr3+(aq)+3e−→Cr(s) −0.73 Cu+(aq)+e−→Cu(s) 0.52 Zn2+(aq)+2e−→Zn(s) −0.76 Cu2+(aq)+2e−→Cu(s) 0.34 Mn2+(aq)+2e−→Mn(s) −1.18 2H+(aq)+2e−→H2(g) 0.00 Al3+(aq)+3e−→Al(s) −1.66 Fe3+(aq)+3e−→Fe(s) −0.036 Mg2+(aq)+2e−→Mg(s) −2.37 Pb2+(aq)+2e−→Pb(s) −0.13 Na+(aq)+e−→Na(s) −2.71 Sn2+(aq)+2e−→Sn(s) −0.14 Ca2+(aq)+2e−→Ca(s) −2.76 Ni2+(aq)+2e−→Ni(s) −0.23 Ba2+(aq)+2e−→Ba(s) −2.90 Co2+(aq)+2e−→Co(s) −0.28 K+(aq)+e−→K(s) −2.92 Cd2+(aq)+2e−→Cd(s)...
Need help with questions 1-5 D Determine whether each redox reaction occurs spontane- ously in the forward direction. (a) Ca2+(aq) + Zn(s)-Ca(s) + Zr"(al) (b) 2 Ag+(aq) + Ni(s)--2 Ag(s) + N产(aq) (c) Fe(s) +Mn2 (aą)- Fe (aą)Mn(s) (d) 2 Al(s) + 3 Pb2+(aq) → 2 AP"(aq) + 3 Pb(s) Suppose you wanted to cause Pb ions to come out of solu- tion as solid Pb. What metal could you use to accomplish this? Make a sketch of an electrochemical...
Using the activity series, which one of the reactions will proceed spontaneously? Beside each, write "WILL PROCEED" (if it will proceed spontaneously) or "WILL NOT PROCEED" (if it will not proceed spontaneously). A) Sn() + Mn2+ (aq) → Sn2+ (aq) + Mn() B) Mg+2 (aq) + Cu(9) Mg ) + Cut2 (aq) C.) 2Ag+ (aq) + Ni () + 2Ag (9) + Ni2+ (a (aq) Given the following balanced net ionic equation: Mn + Cr+2 (aq) + Mn+2 (aq) +...
Half-reaction Cr3+ (aq) + 3e--Cr(s) Fe2+ (aq) + 2e - Fe(s) Fe3+ (aq) - Fe2+ (5) Sn+ (aq) + 2e - Sn2(aq) E (V) -0.74 -0.440 +0.771 +0.154 1. Calculate the standard cell potential for the voltaic cell based on the reaction below, given the table above: 35nt(s) + 2Cr (s) - 2 C (s) + 3 Sn () ANSWER: 2. Calculate the standard cell potential for the voltaic cell based on the reaction below, Riven the table above 3Feb...