how do these things affect eachother? if I have Zn(s)|Zn2+(aq)||Ag+(aq)|Ag(s) and the solution in the salt...
how do these things affect eachother?
if I have Zn(s)|Zn2+(aq)||Ag+(aq)|Ag(s) and the solution in the
salt bridge is changes to CH3OH from potassium nitrate.
if I have Fe3+(aq)|Fe2+(aq)||Ag+(aq)|Ag(s) and the iron (ii) nitrate solution is replaced with salt water
if I have Cu(s)|Cu2+(aq)||MnO4^-(aq)|C(s) and I replaced the carbon electrode with silver
the rest are voltaic cells
if I have Co (s) | Co2+ (aq)|| Cu2+(aq) |Cu (s) instead of a porous cup we turn it into two half cells connected with a pottassium nitrate salt bridge
if I have Sn(s)|Sn2+(aq)||Ag+(aq)|Ag(s) and the tin
electrode is replaced with carbon
also what happens to a voltmeter output when a salt bridge is removed
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how do these things affect eachother?
if I have Zn(s)|Zn2+(aq)||Ag+(aq)|Ag(s) and the solution in the
salt...
Chemistry 30 tUnisn. emistry 30 Unit C Module 6 Summative Assessment Identify how each of the...
Chemistry 30 tUnisn. emistry 30 Unit C Module 6 Summative Assessment Identify how each of the following changes would affect the operation of 11. each cell. Explain each answer for full marks. (5 Marks) Answer Change Voltaic Cell System The solution in the salt-bridge is replaced with CH3OH(aq) There will be a reducsn iA the Cel ooter the sult hciduc used miintaiA eledtica The iron(II) nitrate solution is replaced with salt water ncutr h (NaCl(aq) 1 Result Explanation 2 the...
Half-Reaction Fe2+(aq) + 2e Fe(s) Hg2+ (aq) + 2e Hg() Ag+ (aq) + e + Ag(s)...
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
KNO, salt bridge Mg Zn Ag 1.0 M Mg 1.0 M Zn2 1.0 M Cu 1.0...
KNO, salt bridge Mg Zn Ag 1.0 M Mg 1.0 M Zn2 1.0 M Cu 1.0 M Ag Reset Info 1.0 M ME LO M 2 1.0 MC 10 MA Reset Set up a cell that has a My electrode in a 1.0 M aqueous Mg? solution on the left side and a Ag electrode in a 1.0 M aqueous A solution on the right side. Add the salt bridge. What is the balanced net ionic equation for this reaction?...
In a copper-zinc voltaic cell, one half-cell consists of a ZnZn electrode inserted in a solution...
In a copper-zinc voltaic cell, one half-cell consists of a ZnZn
electrode inserted in a solution of zinc sulfate and the other
half-cell consists of a CuCu electrode inserted in a copper sulfate
solution. These two half-cells are separated by a salt bridge.
At the zinc electrode (anode), ZnZn metal undergoes oxidation by
losing two electrons and enters the solution as Zn2+Zn2+ ions. The
oxidation half-cell reaction that takes place at the anode is
Zn(s)→Zn2+(aq)+2e−Zn(s)→Zn2+(aq)+2e−
The CuCu ions undergo reduction...
salt bridge Zn(s) electrode Culs) electrode 1.0M Zn (a 1.0 M Cu (aq A voltaic cell...
salt bridge Zn(s) electrode Culs) electrode 1.0M Zn (a 1.0 M Cu (aq A voltaic cell similar to that shown in the figure above is constructed. The electronic device shown at the top of the figure is a volt meter. One electrode compartment consists of a zinc strip placed in a 1.0 M ZnCl2 solution, and the other has a copper strip placed in a 1.0 M CuSOA solution. The overall cell reaction is: Zn(s)Cu2+(aq)= zn2+ (aq ) Cu (s)...
Candidate l: Zn(s) | Zn2+(aq,0.500 M) I Cu2+(aq, 1.00 M) Cu(s) Candidate 2: Pb(s) | Pb2+(aq,...
Candidate l: Zn(s) | Zn2+(aq,0.500 M) I Cu2+(aq, 1.00 M) Cu(s) Candidate 2: Pb(s) | Pb2+(aq, 0.500 M) || Cu2+(aq, 1.00 M) Cu(s) Candidate 3: Mg(s) | Mg2+(aq, 0.500 M) | Pb2+(aq, 1.00 M)| Pb(s) (a) 6 pts) Choose one of the candidate voltaic cells #1, #2, or #3. Draw a schematic cell diagram for the candidate voltaic cell of choice. Clearly label anode, cathode, electrodes, ions and their concentrations, salt bridge, and the flow of electrons. (b) (5 pts)...
Calculate the standard free-energy change at 25 ∘C for the following reaction: Mg2+(aq) + Zn(s) → Mg(s) + Zn2+(aq) Expre...
Calculate the standard free-energy change at 25 ∘C for the following reaction: Mg2+(aq) + Zn(s) → Mg(s) + Zn2+(aq) Express your answer to three significant figures and in units of kJ/mol. Consider constructing a voltaic cell with one compartment containing a Zn(s) electrode immersed in a Zn2+ aqueous solution and the other compartment containing an Al(s) electrode immersed in an Al3+ aqueous solution. What is the spontaneous reaction in this cell? Group of answer choices Zn + Al3+ → Al...
The following half-cells are availble: (i) Ag+ (1.0 M) ∣ Ag (s) 0.7994 v (ii) Zn2+...
The following half-cells are availble: (i) Ag+ (1.0 M) ∣ Ag (s) 0.7994 v (ii) Zn2+ (1.0 M) ∣ Zn (s) -0.76 v (iii) Cu2+ (1.0 M) ∣ Cu (s) 0.337 v (iv) Co2+ (1.0 M) ∣ Co (s) -0.28 v Linking any two half-cells makes a voltaic cell. Given four different half-cells, six voltaic cells are possible. These are labeled, for simplicity, Ag - Zn, Ag-Cu, Ag-Co, Zn - Cu, Zn-Co, and Cu-Co. (the first metal is the anode...
The following half-cells are availble: (i) Ag+ (1.0 M) ∣ Ag (s) 0.7994 v (ii) Zn2+...
The following half-cells are availble: (i) Ag+ (1.0 M) ∣ Ag (s) 0.7994 v (ii) Zn2+ (1.0 M) ∣ Zn (s) -0.76 v (iii) Cu2+ (1.0 M) ∣ Cu (s) 0.337 v (iv) Co2+ (1.0 M) ∣ Co (s) -0.28 v Linking any two half-cells makes a voltaic cell. Given four different half-cells, six voltaic cells are possible. These are labeled, for simplicity, Ag - Zn, Ag-Cu, Ag-Co, Zn - Cu, Zn-Co, and Cu-Co. (the first metal is the anode...
Half-cell Potentials: Half Reaction: E value +0.80 V Agt + e → Ag Fe3+ + €...
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...
Chemistry 30 tUnisn. emistry 30 Unit C Module 6 Summative Assessment Identify how each of the following changes would affect the operation of 11. each cell. Explain each answer for full marks. (5 Marks) Answer Change Voltaic Cell System The solution in the salt-bridge is replaced with CH3OH(aq) There will be a reducsn iA the Cel ooter the sult hciduc used miintaiA eledtica The iron(II) nitrate solution is replaced with salt water ncutr h (NaCl(aq) 1 Result Explanation 2 the...
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
KNO, salt bridge Mg Zn Ag 1.0 M Mg 1.0 M Zn2 1.0 M Cu 1.0 M Ag Reset Info 1.0 M ME LO M 2 1.0 MC 10 MA Reset Set up a cell that has a My electrode in a 1.0 M aqueous Mg? solution on the left side and a Ag electrode in a 1.0 M aqueous A solution on the right side. Add the salt bridge. What is the balanced net ionic equation for this reaction?...
In a copper-zinc voltaic cell, one half-cell consists of a ZnZn
electrode inserted in a solution of zinc sulfate and the other
half-cell consists of a CuCu electrode inserted in a copper sulfate
solution. These two half-cells are separated by a salt bridge.
At the zinc electrode (anode), ZnZn metal undergoes oxidation by
losing two electrons and enters the solution as Zn2+Zn2+ ions. The
oxidation half-cell reaction that takes place at the anode is
Zn(s)→Zn2+(aq)+2e−Zn(s)→Zn2+(aq)+2e−
The CuCu ions undergo reduction...
salt bridge Zn(s) electrode Culs) electrode 1.0M Zn (a 1.0 M Cu (aq A voltaic cell similar to that shown in the figure above is constructed. The electronic device shown at the top of the figure is a volt meter. One electrode compartment consists of a zinc strip placed in a 1.0 M ZnCl2 solution, and the other has a copper strip placed in a 1.0 M CuSOA solution. The overall cell reaction is: Zn(s)Cu2+(aq)= zn2+ (aq ) Cu (s)...
Candidate l: Zn(s) | Zn2+(aq,0.500 M) I Cu2+(aq, 1.00 M) Cu(s) Candidate 2: Pb(s) | Pb2+(aq, 0.500 M) || Cu2+(aq, 1.00 M) Cu(s) Candidate 3: Mg(s) | Mg2+(aq, 0.500 M) | Pb2+(aq, 1.00 M)| Pb(s) (a) 6 pts) Choose one of the candidate voltaic cells #1, #2, or #3. Draw a schematic cell diagram for the candidate voltaic cell of choice. Clearly label anode, cathode, electrodes, ions and their concentrations, salt bridge, and the flow of electrons. (b) (5 pts)...
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...
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