# Problem: The standard reduction potentials of the following half-reactions are given in Appendix E in the textbook:l Ag+ (aq) + e-  →  Ag(s)Cu2 +  (aq) + 2e-  →  Cu(s)Ni2 +  (aq) + 2e-  →  Ni(s)Cr3 +  (aq) + 3e-  →  Cr(s).Determine which combination of these half-cell reactions leads to the cell reaction with the smallest positive cell emf.

###### FREE Expert Solution

We’re being asked to determine which combination of these half-cell reactions leads to the cell reaction with the smallest positive cell emf.

Recall that the standard cell potential, E°cell or emf is given by the equation:

$\overline{){\mathbf{E}}{{\mathbf{°}}}_{{\mathbf{cell}}}{\mathbf{=}}{\mathbf{E}}{{\mathbf{°}}}_{{\mathbf{cathode}}}{\mathbf{-}}{\mathbf{E}}{{\mathbf{°}}}_{{\mathbf{anode}}}}$

To do so, we need to do the following:

Step 1. Determine the half-cell potentials (refer to the Standard Reduction Potential Table)
Step 2. Identify the oxidation half-reaction (anode) and the reduction half-reaction (cathode)
Step 3. Calculate E°cell for each combination of half-cell reactions

###### Problem Details
The standard reduction potentials of the following half-reactions are given in Appendix E in the textbook:
l Ag+ (aq) + e-  →  Ag(s)
Cu2 +  (aq) + 2e-  →  Cu(s)
Ni2 +  (aq) + 2e-  →  Ni(s)
Cr3 +  (aq) + 3e-  →  Cr(s).

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