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:

$\overline{){\mathbf{E}}{{\mathbf{\xb0}}}_{{\mathbf{cell}}}{\mathbf{=}}{\mathbf{E}}{{\mathbf{\xb0}}}_{{\mathbf{cathode}}}{\mathbf{-}}{\mathbf{E}}{{\mathbf{\xb0}}}_{{\mathbf{anode}}}}$

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*

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

l Ag^{+} (aq) + e^{-} → Ag(s)

Cu^{2 + } (aq) + 2e^{-} → Cu(s)

Ni^{2 + } (aq) + 2e^{-} → Ni(s)

Cr^{3 + } (aq) + 3e^{-} → Cr(s).

l Ag

Cu

Ni

Cr

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

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