MnO_2(s) + 2H_2 O(I) +2e = Mn^2+ (aq) + 40H^- (aq) epsilon^o = 1.208 v 2HNO_3...
Standard Electrode Potentials at 25?C Reduction Half-Reaction E?(V) F2(g)+2e? ?2F?(aq) 2.87 Au3+(aq)+3e? ?Au(s) 1.50 Cl2(g)+2e? ?2Cl?(aq) 1.36 O2(g)+4H+(aq)+4e? ?2H2O(l) 1.23 Br2(l)+2e? ?2Br?(aq) 1.09 NO3?(aq)+4H+(aq)+3e? ?NO(g)+2H2O(l) 0.96 Ag+(aq)+e? ?Ag(s) 0.80 I2(s)+2e? ?2I?(aq) 0.54 Cu2+(aq)+2e? ?Cu(s) 0.16 2H+(aq)+2e? ?H2(g) 0 Cr3+(aq)+3e? ?Cr(s) -0.73 2H2O(l)+2e? ?H2(g)+2OH?(aq) -0.83 Mn2+(aq)+2e? ?Mn(s) -1.18 How can the table be used to predict whether or not a metal will dissolve in HCl? In HNO3? Drag the terms on the left to the appropriate blanks on the right to...
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...
Using standard reduction potentials from the ALEKS Data tab, calculate the standard reaction free energy AG for the following redox reaction. Round your answer to 3 significant digits Mn (aq)+2H,O()+2Fe (aq)MnO, ()+4H (aq)+2Fe (aq) dhData Cu (aq) + e Cu (5) F2 (0)+2e2F (aq) Fe (aq) +2e Fe (s) Fe (aq) + eFe2 (aq) Fe (aq) + 3e Fe (s) 2.866 X ? -0.447 0.771 -e.037 2H (aq)+2e H (0) e.000 2H)O (I)+2e H2 (a) +20H(aq) -0.8277 1.776 H2O2 (aq)...
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...
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)...
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Half-reaction E° (V) Br26) + 2e —> 2Br (aq) 1.080V Ca2+(aq) + 2e —— Cd) 0.403V Zn2+ (aq) + 2e —— Zn(0.763V (1) The strongest oxidizing agent is: enter formula (2) The weakest oxidizing agent is: (3) The weakest reducing agent is: (4) The strongest reducing agent is: (5) Will Br2(1) oxidize Zn(s) to Zn?"(aq)? O (6) Which species can be oxidized by Ca2+(aq)? If none, leave box blank. Consider the following half-reactions: Half-reaction E° (V)...
Given the following standard reduction potentials: H^+ (aq) + 2e^- rightarrow H_2(g) E degree = 0.00 V Fe^3+ (aq) + 2e^- rightarrow Fe(s) E degree = -0.43 V (a) What is the cell potential by combining the above two half-reactions to make a working voltaic cell (same as galvanic cell)? (b) Which species will be oxidized in anode? Write the half-reaction for the anode. (c) Write the overall reaction and balance the chemical equation for this working voltaic cell. (d)...
Consider the following half-reactions: Half-reaction E° (V) 12(s) + 2e - →21"(aq) 0.535V 2H+ (aq) + 2e - → H2(g) 0.000V Cr3+(aq) + 3e —— Cr(s) -0.740V The strongest oxidizing agent is: enter formula The weakest oxidizing agent is: The weakest reducing agent is: The strongest reducing agent is: Will 12(s) reduce Cr3+(aq) to Cr(s)? — Which species can be reduced by H2(g)? If none, leave box blank. Use the References to access important values if needed for this question....
Standard Reduction Please write your answers here Reduction Half-Reactin Potential (V) F2(g) + 2e-→ 2F-(aq) S2082 (ag) +2e-2SO42(ag) O2(g) + 4H(a)+ 4e 2H200) +2.87 +2.01 +1.23 +1.09 +0.80 +0.77 +0.54 +0.34 +0.15 +0.14 0.00 0.14 0.26 0.44 0.74 0.76 0.83 1.18 2.71 3.04 2 4 Ag+(aq) + e-→ Ag(s) Fe3+(ag)eFe2*(aq) 20)+ 2e- 21(aq) Cu2(ag)+ 2e Cus) SAMPLE QUIZ 4 S(s) + 2H+(aq) + 2e. → H2S(g) 2H(a)+ 2eH2g) Sn2(ag) 2e Sng) 1. What is the purpose of the salt bridge...
12. Using two half reactions that have NEGATIVE standard reduction potentials results results in a battery that... Reduction Half-Reaction F2(g) + 2e →2F(aq) S2082 (aq) + 2e- → 25042 (aq) O2(g) + 4H+ (aq) + 4e → 2H2O(1) Br2(1) + 2e + 2Br (aq) Agt(aq) + e → Ag(s) Fe3+ (aq) + e- → Fe2+ (aq) 126) + 2e → 21 (aq) Cu2+ (aq) + 2e → Cu(s) Sn4+ (aq) + 2e → Sn2+ (aq) S(s) + 2H+ (aq) +...