1860 United States presidential election in Missouri

1860 United States presidential election in Missouri

← 1856 November 2, 1860 1864 →

Nominee Stephen A. Douglas John Bell Party Democratic Constitutional Union Home state Illinois Tennessee Running mate Herschel Vespasian Johnson Edward Everett Electoral vote 9 0 Popular vote 58,801 58,372 Percentage 35.52% 35.26%   Nominee John C. Breckinridge Abraham Lincoln Party Southern Democratic Republican Home state Kentucky Illinois Running mate Joseph Lane Hannibal Hamlin Electoral vote 0 0 Popular vote 31,362 17,028 Percentage 18.94% 10.28%

County Results[a] Douglas   40-50%   50-60%   60-70% Bell   40-50%   50-60%   60-70% Breckinridge   40-50%   50-60%   60-70%   80-90% Lincoln   40-50%   50-60% Unknown/No Vote

President before election James Buchanan Democratic Elected President Abraham Lincoln Republican

Elections in Missouri
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The 1860 United States presidential election in Missouri took place on November 2, 1860, as part of the 1860 United States presidential election. Voters chose nine representatives, or electors to the Electoral College, who voted for president and vice president.

Missouri was won by Democratic candidate, Stephen A. Douglas. He won the state by a very narrow margin of 0.26%. The state was the only one to fully give its votes to Douglas, though he would win the popular vote and three of the seven electoral votes from New Jersey under a fusion ticket.

As of the 2020 presidential election, this is the last occasion when Putnam County voted for a Democratic presidential candidate and the last occasion when Taney County did not vote for the Republican candidate.^[1]

Results[]

See also[]

• United States presidential elections in Missouri

Notes[]

References[]

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