1860 United States presidential election in Iowa

Summary

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

← 1856 November 6, 1860 1864 →
 
Nominee Abraham Lincoln Stephen A. Douglas
Party Republican Democratic
Home state Illinois Illinois
Running mate Hannibal Hamlin Herschel V. Johnson
Electoral vote 4 0
Popular vote 70,302 55,639
Percentage 54.61% 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 edit

1860 United States presidential election in Iowa[1]
Party Candidate Votes %
Republican Abraham Lincoln 70,302 54.61%
Democratic Stephen A. Douglas 55,639 43.22%
Constitutional Union John Bell 1,763 1.37%
Southern Democratic John C. Breckinridge 1,035 0.80%
Total votes 128,739 100%

See also edit

References edit

  1. ^ "1860 Presidential Election Results Iowa".