The 1860 United States presidential election in Minnesota took place on November 6, 1860, as part of the 1860 United States presidential election. Minnesota voters chose four representatives, or electors, to the Electoral College, who voted for president and vice president.
Minnesota voted in its first ever United States presidential election, having been admitted as the 32nd state on May 11, 1858. The state was won by IllinoisRepresentativeAbraham Lincoln (Republican Party (United States)), running with SenatorHannibal Hamlin, with 63.44% of the popular vote, against SenatorStephen A. Douglas (D–Illinois), running with 41st Governor of GeorgiaHerschel V. Johnson, with 34.27% of the popular vote.
With 63.44% of the popular vote, Lincoln's victory within the state would be his second strongest victory in terms of percentage in the popular vote in the 1860 election after Vermont.[1] Minnesota would never vote Democratic until Franklin D. Roosevelt won it in 1932, 72 years later.
Results[]
1860 United States presidential election in Minnesota[2]
Party
Candidate
Votes
%
Republican
Abraham Lincoln
22,069
63.44%
Democratic
Stephen A. Douglas
11,920
34.27%
Southern Democratic
John C. Breckinridge
748
2.15%
Constitutional Union
John Bell
50
0.14%
Total votes
34,787
100%
v
t
State results of the 1860 U.S. presidential election