Can someone explain the anticlockwise rule for electrode potentials?

The anticlockwise rule is a quick way to determine whether a redox reaction will occur between two species.

Here's how it works.

Example

Will metallic #"Ag"# react with an aqueous solution of #"Cu"^"2+"#?

Solution

Step 1. Write the standard reduction half-reactions for #"Ag"# and #"Cu"#

Make sure that you write the half-reaction with the more negative ( less positive) potential first. This gives

#"Cu"^"2+""(aq)" + 2"e"^"-" ⇌ "Cu(s)"; E^@ = "+0.34 V"#
#"Ag"^"+""(aq)" + "e"^"-" ⇌ "Ag(s)"; color(white)(m)E^@ = "+0.80 V"#

Step 2. Use the anticlockwise rule

Draw anticlockwise curved arrows (the blue arrows in the diagram below) anticlockwise above and below your equations.

The #bb(color(blue)("anticlockwise rule")# states that the half-reactions that will occur are those that follow anticlockwise paths.

Step 3. Determine the equation for the reaction that will occur

We must reverse the top equation.

#"Cu(s)" ⇌ "Cu"^"2+""(aq)" + 2"e"^"-" ;color(white)(mmmmmml) E^@ =color(white)(l) "-0.34 V"#
#ul("2Ag"^"+""(aq)" + 2"e"^"-" ⇌ "2Ag(s)";color(white)(mmmmmll) E^@ = "+0.80 V")#
#"Cu(s) + 2Ag"^"+" ⇌ "Cu"^"2+""(aq)" + "2Ag(s)";color(white)(l) E^@ = "+0.46 V"#

The rule predicts that #"Cu"# will react with #"Ag"^"+""# to form #"Ag"# and #"Cu"^"2+"#.

The other conclusion is that #"Ag"# will not react with #"Cu"^"2+"#.