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The standard reduction potentials of the following half-reactions are given in Appendix E in the textbook:...

The standard reduction potentials of the following half-reactions are given in Appendix E in the textbook:

Ag+(aq)+e−→Ag(s)= .799

Cu2+(aq)+2e−→Cu(s)= .337

Ni2+(aq)+2e−→Ni(s)= -.28

Cr3+(aq)+3e−→Cr(s). = -.74

1. Determine which combination of these half-cell reactions leads to the cell reaction with the largest positive cell emf.

1st and 2nd,
1st and 3rd,
1st and 4th,
2nd and 3rd,
3rd and 4th.

It isn't the first or last one because I have gotten it wrong twice.

2. Calculate the value of this emf.

3. Then determine which combination is the smallest and calculate the emf.

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Answer #1

1] Ag+(aq)+e−→Ag(s)= 0.799

Cr3+(aq)+3e−→Cr(s). = -0.74

The combination 1st and 4th cell provides large psoitive emf

2] 3Ag+ + Cr -----> 3Ag + Cr+3

The emf of the reaction is 0.799+0.74 = 1.539 V

3] 3rd and 4th combination gives small postive emf

Ni2+(aq)+2e−→Ni(s)

Cr3+(aq)+3e−→Cr(s)

----------------------------------

3Ni+2 + 2Cr------> 3Ni + 2Cr+3

Emf = 0.74 - 0.28 = 0.46 V

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