1860 United States presidential election in Iowa

The 1860 United States presidential election in Iowa took place on November 6, 1860, as part of the 1860 United States presidential election. Iowa voters chose four representatives, or electors, to the Electoral College, who voted for president and vice president.

1860 United States presidential election in Iowa

← 1856November 6, 18601864 →
 
NomineeAbraham LincolnStephen A. Douglas
PartyRepublicanDemocratic
Home stateIllinoisIllinois
Running mateHannibal HamlinHerschel V. Johnson
Electoral vote40
Popular vote70,30255,639
Percentage54.61%43.22%

County Results

President before election

James Buchanan
Democratic

Elected President

Abraham Lincoln
Republican

Iowa was won by the Republican nominees Illinois Representative Abraham Lincoln and his running mate Senator Hannibal Hamlin of Maine. They defeated the Democratic nominees Senator Stephen A. Douglas of Illinois and his running mate 41st Governor of Georgia Herschel V. Johnson. Lincoln won the state by a margin of 11.39%.

Results

1860 United States presidential election in Iowa[1]
PartyCandidateVotes%
RepublicanAbraham Lincoln 70,302 54.61%
DemocraticStephen A. Douglas55,63943.22%
Constitutional UnionJohn Bell1,7631.37%
Southern DemocraticJohn C. Breckinridge1,0350.80%
Total votes128,739 100%

See also

References