1836 United States presidential election in New Jersey

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United States presidential election in New Jersey, 1836

← 1832 November 3 - December 7, 1836 1840 →
  Van Buren.jpg William Henry Harrison by James Reid Lambdin, 1835 crop.jpg
Nominee Martin Van Buren William Henry Harrison
Party Democratic Whig
Home state New York Ohio
Running mate Richard Mentor Johnson Francis Granger
Electoral vote 8 0
Popular vote 26,137 25,592
Percentage 50.53% 49.47%

President before election

Andrew Jackson
Democratic

Elected President

Martin Van Buren
Democratic

The 1836 United States presidential election in New Jersey took place between November 3 and December 7, 1836, as part of the 1836 United States presidential election. Voters chose eight representatives, or electors to the Electoral College, who voted for President and Vice President.

New Jersey voted for the Democratic candidate, Martin Van Buren, over Whig candidate William Henry Harrison. Van Buren won New Jersey by a margin of 1.06%.

Results[edit]

United States presidential election in New Jersey, 1836[1]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Whig William Henry Harrison of Ohio Francis Granger of New York 26,137 50.53% 8 100.00%
Democratic Martin Van Buren of New York Richard M. Johnson of Kentucky 25,592 49.47% 0 0.00%
Total 51,729 100.00% 8 100.00%

References[edit]

  1. ^ "1836 Presidential General Election Results - New Jersey". U.S. Election Atlas. Retrieved 23 December 2013.