Answer 2
Row 1 Reaction with Ag
2Ag+ + H2 ------> 2Ag + 2H+
Reduction of silver by hydrogen.
Row 2 Reaction with Al
2Al + 6H+ ------> 2Al3+ + 3H2
Oxidation of Aluminum by hydrogen ion.
Row 3 Reaction with Fe
Fe + 2H+ ------> Fe2+ + H2
Oxidation of iron by hydrogen ion.
Row 4 Reaction with Zn
Zn + 2H+ ------> Zn2+ + H2
Oxidation of zinc by hydrogen ion.
Answer 3
Best Agreement : Silver (Ag+/Ag)
Poorest agreement : Aluminum (Al3+/Al)
Number line resembles for order as well as values except for Al and Fe.
Place for Fe is correct but its value is incorrect.
Actual place for Aluminum should be at the lowest point i.e. below Zinc.
Answer 4
If the cell potential is negative,
The spontaneity of the reaction : non-spontaneous
Reverse of the reaction : spontaneous (cell potential will be positive for reverse reaction).
Chemistry: Experimental Cell Potential Lab report please help checking my work and answering question #2 #3 #4 Thanks!...
Question 4 Using the table of standard reduction potentials shown below, calculate the standard cell potential for a battery based on the following reactions. • MnO4 +8H+ + 5e + Mn+2 +4 H20 . 5 Ag + 5 Ag+1 +5e Reduction Half-Reaction F2 +2e + 2F MnO4 + 8 H+ + 5e + Mn+2 + 4H2O Cl2 + 2e + 20 O2 + 4H+ + 4e + 2 H2O Ag++ e + Ag Fet3 Fe + Fet2 O2 + 2...
Question 7 Using the table of standard reduction potentials shown below, calculate the standard cell potential for a battery based on the following reactions. • Pet2 + 2e + Fe . 2 Li + 2 Li + 2e Reduction Half-Reaction F2 +2e + 2F MnO + 8H+ Se + Mn+2+ 4 H 0 Cl2 + 2e + 20 02 + 4H' + 4e + 2 H2O Ag+ e -- Ag Fet) + e + Fe2 O2 + 2 H2O +...
Question 5 Using the table of standard reduction potentials shown below, calculate the standard cell potential for a battery based on the following reactions. • O2 + 4H+ + 4e + 2 H20 • 2 Cu + 2 Cu+2 + 4e Reduction Half-Reaction F2 + 2e + 2F MnO, +8 H+ + 5e + Mn+2 +4 H20 Cl2 + 2e → 2C O2 + 4H+ + 4e + 2 H2O Agt! + e + Ag Fet3 + e - Fet2...
need help for half cell potentials pls calculate step by step (NOTE - Remember that the positive electrode is attached to the red wire and the negative electrode is attached to the black wire.) Electrode Systems Used Anode (oxidation) Negative Cathode (reduction) Positive Measured Potential (V) Positive (Ecu) Copper and silver Cu() - Cu2+ + 2e" Ag+ +1e Ag) 0.432 V Zinc + Silver Zn cs + 2n**+ Zé dat + leº nAg (s) 1.484 V Copper & Zinc Zn(s)...
need help with the rest of the table EXPERIMENT 10 DETERMINATION OF THE ELECTROCHEMICAL SERIES PURPOSE To determine the standard cell potential values of several electrochemical coll INTRODUCTION The basis for an electrochemical cell is an oxidation reduction Corredor be divided into two half reactions reaction. This reaction can Oxidation half reaction Gloss of electrons) takes place at the anode, which is the positive electrode that the anions migrate to Chence the name anode) Reduction half reaction (gain of electrons)...
Calculate the theoretical cell potential (E°) of a galvanic cell under standard conditions made up of copper and magnesium (see Part II and Table 1 for more information). PARTIL Creating and Testing Voltaic Cells Introduction and Background for the Voltaic Cells A galvanic cell (sometimes more appropriately called a voltaic cell) consists of two half-cells joined by a salt bridge that allow ions to pass between the two sides in order to maintain electroneutrality. Each half-cell contains the Components of...
Using the table below: 19. Three combinations of metals are listed below, which combination would produce the largest voltage if they were used to construct an electrochemical cell? Copper (Cu) with zinc (Zn) Lead (Pb) with zinc (Zn) Lead (Pb) with cadmium (Cd) Liu lur the reaction between Zn and Cu2+ ions is 1.1030 V, we can use the known value for the half-cell potential for zinc to determine the half-cell potential for copper: Zn(s) → Zn2+(aq) + 2e +...
Write the half reactions and overall reaction for each cell with calculated overall potentials as shown in Table 5-1. (Note: for the iron solutions the Nernst equation must be used) Pb(s) | Pb(NO3)2 (0.1M) || Cu(NO3)2 (0.1M) Cu(s) Zn(s) | Zn(NO3)2(0.1M) || Cu(NO3)2 (0.1M) Cu(s) Cds) | Ca(NO3)2 (0.1M) || Cu(NO3)2 (0.1M) | Cu(s) Cu() Cu(NO3)2(0.1M) Il Fe (0.1M/Fe? (0.1M graphite Pb(s) Pb(NO3)2(0.1M) Il Fe3(aq) (0.1M)/ Fe2(aq) (0.1MI graphite(s) Zns | Zn(NO3)2 (0.1M) || Pb(NO3)2 (0.1M) | Pb(s) Cdis Ca(NO3)2...
Using the information in the table: Which combination of metals, if used to create an electrochemical cell, would produce the largest voltage? Liu lur the reaction between Zn and Cu2+ ions is 1.1030 V, we can use the known value for the half-cell potential for zinc to determine the half-cell potential for copper: Zn(s) → Zn2+(aq) + 2e + Cu2+(aq) + 2e → Cu(s) Zn(s) + Cu2+ (aq) → Zn2+ (aq) + Cu(s) E half-cell = 0.7628 V Eºhalf-cell =...
I need help with questione 1-12 and discussion question 1 and 2. The previous pictures help determine the chart. Please Show Work thank you so much An oxidation half-reaction is characterized by electrons appearing on the product side. The oxidation of aluminum for instance would be represented thusly: Al(s) → Al3+ + 3e- (1) An reduction half-reaction is characterized by electrons appearing on the reactant side. The reduction of ferrous iron for instance would be represented thusly: Fe2+ + 2e...