1860 United States presidential election in Iowa

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1860 United States presidential election in Iowa
Flag of Iowa.svg
  1856 November 6, 1860 1864  
  Abraham Lincoln by Alexander Hesler, 1860-restored (cropped).png Senator Stephen A. Douglas (edited).png
Nominee Abraham Lincoln Stephen A. Douglas
Party Republican Democratic
Home state Illinois Illinois
Running mate Hannibal Hamlin Herschel V. Johnson
Electoral vote40
Popular vote70,30255,639
Percentage54.61%43.22%

Iowa Presidential Election Results 1860.svg
County Results

President before election

James Buchanan
Democratic

Elected President

Abraham Lincoln
Republican

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.

Contents

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%
Republican Abraham Lincoln 70,302 54.61%
Democratic Stephen A. Douglas 55,63943.22%
Constitutional Union John Bell 1,7631.37%
Southern Democratic John C. Breckinridge 1,0350.80%
Total votes128,739 100%

See also

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References

  1. "1860 Presidential Election Results Iowa".