Q. Consider a galvanic cell with a zinc electrode immersed in 1.0M Zn2+ and a silver...
Q. Consider a galvanic cell with a zinc electrode immersed in 1.0M Zn2+ and a silver electrode immersed in 1.0M Ag+. Which of the electrodes is the anode?
Zn2 + 2e- --> Zn E° = -0.76 V
Ag+ + e- --> Ag E° = 0.80 V
Homework Answers
The one with greater Eo value is reduced.
So, that becomes cathode
The one with smaller Eo value is oxidised.
So, that becomes anode.
Eo is smaller for Zn2+ + 2e- -> Zn
So, this is anode
Answer: zinc electrode is anode
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Q. Consider a galvanic cell with a zinc electrode immersed in
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Consider an electrochemical cell with a zinc electrode immersed in 1.0 M Zn2and a silver electrode immersed in 1.0 M Agt. Zn2+ + 2e- → Zn E° = -0.76 V Agt + e- Ag E = 0.80 V Calculate Eº and AGº for this cell Calculate AGº for the reaction: 2 Zn (g) + O2 (g) + 2 H20 (1) ► 2 Zn2+ (aq) + 4 OH (aq) Reduction Half-Reaction E° (V) O2(g) + 2 H20 (1)+4 e-®4 OH(aq) +0.403...
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A galvanic cell is constructed using a zinc electrode immersed in a 1.0 M Zn2+ solution and a silver electrode immersed in a 1.0 M Ag+ solution. What is the overall equation for this cell?
For a voltaic (or galvanic) cell using Ag, Ag+ (1.0M) and Zn Zn2+ (1.0 M) half-cells, which of the following statements is incorrect
For a voltaic (or galvanic) cell using Ag, Ag+ (1.0M) and Zn Zn2+ (1.0 M) half-cells, which of the following statements is incorrect? (a) The zinc electrode is the anode. (b) Electrons will flow through the external circuit from the zinc electrode to the silver electrode. (c) The mass of the zinc electrode will decrease as the cell operates. (d) The concentration of Ag will decrease as the cell operates. (e) Reduction occurs at the zinc electrode as the cell operates.
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Instead of using zinc, a student accidentally used a 1 M solution of silver (Ag). Based on the information in the table below, what would be the theoretical cell voltage, Eocell, when it is used in a galvanic cell with copper? Electrode Eo Ag+ + e- à Ag +0.80 V Cu2+ + 2e- à Cu +0.34 V Pb2+ + 2e- à Pb -0.13 V Zn2+ + 2e- à Zn -0.76 V Al3+ + 3e- à Al -1.66 V
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please solve all correctly Question: Consider the voltaic cell using silver and zinc. The net equation...
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Question: Consider the voltaic cell using silver and zinc. The net equation for this voltaic cell is 2Ag+ (aq) + Zn(s)--> 2Ag(s) + Zn2+lag) Calculate the molar concentration of Ag+ if a potential of 1.58 V was measured across the cell when the molar concentration of Zn* was 0.150 M. Hint: You need to first calculate the of the cell. Given the following Data: Be sure to show all calculations Half-Reaction Standard Potential E* (volts) Ag...
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Information for Questions 8 -9: A concentration cell is set up with two different Zn (s) / Zn2+ (aq) half-cells that contain different concentrations of Zn2+ (aq). The two half-cell solutions are connected with a salt bridge, and the zinc electrodes are connected with a wire and voltmeter. Consider the following standard reduction potential at 25°C: Zn2+ (aq) -0.76 V 2e--> Zn (s) E。= + Use the information here about this concentration cell for Questions 8 and 9. 8. What...
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Consider an electrochemical cell with a zinc electrode immersed in 1.0 M Zn2and a silver electrode immersed in 1.0 M Agt. Zn2+ + 2e- → Zn E° = -0.76 V Agt + e- Ag E = 0.80 V Calculate Eº and AGº for this cell Calculate AGº for the reaction: 2 Zn (g) + O2 (g) + 2 H20 (1) ► 2 Zn2+ (aq) + 4 OH (aq) Reduction Half-Reaction E° (V) O2(g) + 2 H20 (1)+4 e-®4 OH(aq) +0.403...
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