The standard reduction potentials of the following half-reactions are given in A

Question

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

Ag+(aq)+eAg(s)= .799

Cu2+(aq)+2eCu(s)= .337

Ni2+(aq)+2eNi(s)= -.28

Cr3+(aq)+3eCr(s). = -.74

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

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.

Explanation / Answer

1] Ag+(aq)+eAg(s)= 0.799

Cr3+(aq)+3eCr(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)+2eNi(s)

Cr3+(aq)+3eCr(s)

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

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

Emf = 0.74 - 0.28 = 0.46 V

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