The anode in a voltaic cell and in an electrolytic cell is positive in both cell

Question

The anode in a voltaic cell and in an electrolytic cell is positive in both cells the site of oxidation and of reduction, respectively the site of reduction and of oxidation, respectively the site of oxidation in both cells the site of reduction in both cells A strip of copper is placed in a 1 M solution of copper nitrate and a strip of silver is placed in a 1 M solution of silver nitrate. The two metal strips are connected to a voltmeter by wires and a salt bridge connects the solutions. The following standard reduction potentials apply: Which of the following statements is false? Electrons flow in the external circuit from the copper electrode to the silver electrode. The silver electrode increases in mass as the cell operates. There is a net general movement of silver ions through the salt bridge to the copper half-cell. Negative ions pass through the salt bridge from the silver half-cell to the copper Some live copper ions pass through the salt bridge from the copper half-cell to the silver half-cell. which of the following species cannot function as an oxidizing agent.

Explanation / Answer

5. Remember that anode in any cell is defined as the place where the oxidation occurs: correct option D

6. As a voltmeter is connected to our solution, we have to think that this is a voltaic cell (spontaneus reaction), so we are looking to get a positive E°:

Our reaction would be Cu + 2Ag+ ----> Cu2+ + 2Ag in this way the Cu use -0.34 V (as it is inverted) and Ag use +1.60 V (as it is multipied by a factor of 2): Our cell diagram would be: Cu(s)|Cu2+(aq)||Ag+(aq)|Ag(s)

So, seing your options:

A. As Cu is in the process of oxidation: is losing electrons, and the silver is receiving them, TRUE

B. As silver is converting to silver solid, it has to gain mass TRUE

C. False

D. True

E. True:

7. The one that cannot function as an oxidizing agent is the one that cannot gain any more electrons: the only one that has this feature is: OPTION D, iodide can't because it is already in its lowest oxidation state of -1. As an oxidising agent it would have to be a able to lower its oxidation state to -2 (or lower) which is impossible because it would have to expand it´s octet to 9. The other ones are good oxidising agents.

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