1836 United States presidential election in Vermont

Last updated

1836 United States presidential election in Vermont
Flag of Vermont.svg
  1832 November 15, 1836 1840  
  MA
NC  
  William Henry Harrison by James Reid Lambdin, 1835 crop.jpg Martin Van Buren circa 1837 crop.jpg
Nominee William Henry Harrison Martin Van Buren
Party Whig Democratic
Home state Ohio New York
Running mate Francis Granger Richard M. Johnson
Electoral vote70
Popular vote20,99414,037
Percentage59.93%40.07%

Vermont Presidential Election Results 1836.svg
County results

President before election

Andrew Jackson
Democratic

Elected President

Martin Van Buren
Democratic

A presidential election was held in Vermont on November 15, 1836 as part of the 1836 United States presidential election. [1] Voters chose seven representatives, or electors to the Electoral College, who voted for President and Vice President.

Contents

Vermont voted for Whig candidate William Henry Harrison over Democratic candidate Martin Van Buren. Harrison won Vermont by a margin of 19.86%.

This would be the final time a Democratic candidate would carry Essex County until Franklin D. Roosevelt won it 104 years later in 1940.

1836 would stand as the strongest performance for a Democratic candidate in Vermont until 96 years later in 1932, when Franklin D. Roosevelt performed slightly better with 41.08%.

Harrison would later win Vermont again four years later when he defeated Van Buren.

Results

1836 United States presidential election in Vermont [2]
PartyCandidateRunning matePopular voteElectoral vote
Count%Count%
Whig William Henry Harrison of Ohio Francis Granger of New York 20,99459.93%7100.00%
Democratic Martin Van Buren of New York Richard M. Johnson of Kentucky 14,03740.07%00.00%
Total35,031100.00%7100.00%

See also

References

  1. "Presidential Elections". Weekly Messenger. November 12, 1836.
  2. "1836 Presidential General Election Results - Vermont". U.S. Election Atlas. Retrieved December 23, 2013.