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Using the standard electrode potentials Co2+(aq) + 2e → Co(s) E° = -0.28 V Fe2+(aq) +...
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...
Consider these two equations. Co2 + (aq) + 2e--> Co(s) Standard reduction potentials can be found here. What is the standard potential of a cell where magnesium is the anode and cobalt is the cathode? O -2.66 V O -2.10 V O 2.66 V O 2.10 V
Half-cell Potentials: Half Reaction: E value +0.80 V Agt + e → Ag Fe3+ + € → Fe2+ +0.77 v +0.34 V -0.13 V Cu2+ +2e → Cu Pb2+ + 2e - → Ib Ni2+ + 2e → Ni Cd2+ +2e → Cd -0.25 V -0.40 V Fe2+ + 2e → Fe -0.44 V Zn2+ + 2e → Zn -0.76 V Al3+ +3e → AI - 1.66 V Consider an electrochemical cell constructed from the following half cells, linked by...
Two half-reactions are shown with their standard cell potentials. If a galvanic cell is constructed using them, which electrode would be the anode, and what would the cell potential be? Fe2+ (aq) + 2e → Fe(s); E=-0.44 V Ag+(aq) + 1e → Ag(s); E=0.80 V O a. Iron, 1.24 V O b. Silver, 0.36 V O c. Iron, 0.36 V O d. Iron, -0.36 V O e. Silver, -0.36 V
+ Given the following electrode potentials at 25°C Fe3+ e-- Fe2+ E° = 0.571 V 2e Fe(s) E° = -0.440 V Calculate the electrode potential for Fe3+ + 3 e- Fe(s) Fe2+ + Select one: a. -0.132 b. -0.036 c. 0.081 d.-0.211 e. 0.103
Half-Reaction Fe2+(aq) + 2e Fe(s) Hg2+ (aq) + 2e Hg() Ag+ (aq) + e + Ag(s) Cu2+ (aq) + 2e + Cu(s) Zn2+ (aq) + 2e → Zn(s) E (V) -0.44 0.86 0.80 0.34 - 0.76 Using the table, calculate Eºcell for the following electrochemical cell under standard conditions voltmeter a) 1.24 V Fe. salt bridge Ag b) -1.24 V c) 2.04 V d) - 2.04 V Ag a b С
19 20 Question 16 Half-cell Potentials: Half Reaction: E' value + 0.80 V +0.77 V Agte → AS Fe3+ + + Fe2+ Cu2+ 2e → Cu Pb2+ + 2e → Pb +0.34 V -0.13 V Ni2+ + 2e → NI -0.25 V - 0.40 V Cd2+ +2e → ca Fe2+ + 2e → Fe Zn2+ + 2e → Zn -0.44 V - 0.76 V A13+ +3 → AI - 1.66 V Consider an electrochemical cell constructed from the following half...
The following data were measured using a nickel electrode as the standard: Cu2+(aq) + 2e- ⟶ Cu(s) Ni2+(aq) + 2e- ⟶ Ni(s) Fe2+(aq) + 2e- ⟶ Fe(s) Al3+(aq) + 3e- ⟶ Al(s) The copper and aluminum electrodes are connected in a battery. A) Which is the anode? Cathode? Why? B) Which is oxidized? Reduced? Why? C) What will the battery voltage be? Eo =0.62V Eo =0.00V Eo =-0.15V Eo =-1.38V D) Write a balanced net ionic equation for the reaction...
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...
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...