If a voltaic cell is constructed consisting of a Cu/Cu 2+ half cell and a Ag/Ag

Question

If a voltaic cell is constructed consisting of a Cu/Cu2+ half cell and a Ag/Ag+ half cell under standard conditions, what reaction should occur at the anode? You will need to refer to the table of reduction potentials in Appendix E in your textbook.

Cu(s) Cu2+(aq) +   2e

b.) Ag(s) Ag+(aq) + e

c.) Cu2+(aq)   + 2e Cu(s)

d.) Ag+(aq) + e Ag(s)

any help/ explinaiton would be appreciated :) thanks

Cu(s) Cu2+(aq) +   2e

b.) Ag(s) Ag+(aq) + e

c.) Cu2+(aq)   + 2e Cu(s)

d.) Ag+(aq) + e Ag(s)

any help/ explinaiton would be appreciated :) thanks

Explanation / Answer

In voltaic cell, at an anode oxidation reaction takes place and at cathode reduction reaction takes place.

An oxidizing agent with a higher reduction potential will oxidize (and itself get reduced) any reducing agent with a lower reduction potential.

A reducing agent with a lower reduction potential will reduce (and itself get oxidized) any oxidizing agent with a higher reduction potential.

Standard reduction electrode potentials for,

Cu/Cu2+ is +0.340 V

Ag/Ag+ is + 0.800 V.

It’s clear that Standard reduction electrode potential for Cu/Cu2+ is lower than that of Ag/Ag+ and hence, Cu will reduce Ag+ ---- > Ag and itself get oxidized to Cu2+.

i.e. Cu(s) --------- > Cu2+ + 2e- this oxidation reaction will occur at an anode. Option-(a).

And at cathode, Ag+ (aq.) + e- ------ > Ag(s) this reduction reaction will occur.

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