1856 United States presidential election in New Jersey

The 1856 United States presidential election in New Jersey took place on November 4, 1856, as part of the 1856 United States presidential election. Voters chose seven representatives, or electors to the Electoral College, who voted for president and vice president.

1856 United States presidential election in New Jersey

November 4, 1856
 
Nominee James Buchanan John C. Frémont Millard Fillmore
Party Democratic Republican Know Nothing
Home state Pennsylvania California New York
Running mate John C. Breckinridge William L. Dayton Andrew Jackson Donelson
Electoral vote 7 0 0
Popular vote 46,943 28,338 24,115
Percentage 47.23% 28.51% 24.26%

President before election

Franklin Pierce
Democratic

Elected President

James Buchanan
Democratic

New Jersey voted for the Democratic candidate, James Buchanan, over Republican candidate, John C. Frémont, and the Know Nothing candidate, Millard Fillmore. Buchanan won the state by a margin of 13.43 percentage points, and would be the last Democratic presidential candidate to carry Cumberland County until Franklin D. Roosevelt in 1936, and the last to carry Passaic County until Roosevelt in 1932.[1]

Results

1856 United States presidential election in New Jersey[2]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Democratic James Buchanan of Pennsylvania John C. Breckinridge of Kentucky 46,943 47.23% 7 100.00%
Republican John C. Frémont of California William L. Dayton of New Jersey 28,338 28.51% 0 0.00%
Know Nothing Millard Fillmore of New York Andrew Jackson Donelson of Tennessee 24,115 24.26% 0 0.00%
Total 99,396 100.00% 7 100.00%

See also

References

  1. Menendez, Albert J.; The Geography of Presidential Elections in the United States, 1868-2004, pp. 258-259 ISBN 0786422173
  2. "1856 Presidential General Election Results – New Jersey".
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