1836 United States presidential election in New Jersey
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 8 representatives, or electors to the Electoral College, who voted for President and Vice President.
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Elections in New Jersey | ||||||||
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New Jersey voted for the Whig candidate, William Henry Harrison, over Democratic candidate Martin Van Buren. Harrison won New Jersey by a margin of 1.06%.
Results
1836 United States presidential election in New Jersey[1] | ||||||||
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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% |
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References
- "1836 Presidential General Election Results - New Jersey". U.S. Election Atlas. Retrieved 23 December 2013.
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