1836 United States presidential election in Vermont explained

See main article: 1836 United States presidential election.

Election Name:1836 United States presidential election in Vermont
Country:Vermont
Flag Year:1804
Type:presidential
Ongoing:no
Previous Election:1832 United States presidential election in Vermont
Previous Year:1832
Next Election:1840 United States presidential election in Vermont
Next Year:1840
Election Date:November 3 - December 7, 1836
Image1:William Henry Harrison by James Reid Lambdin, 1835 crop.jpg
Nominee1:William Henry Harrison
Party1:Whig Party (United States)
Home State1:Ohio
Running Mate1:Francis Granger
Electoral Vote1:7
Popular Vote1:20,994
Percentage1:59.93%
Nominee2:Martin Van Buren
Party2:Democratic Party (United States)
Home State2:New York
Running Mate2:Richard Mentor Johnson
Electoral Vote2:0
Popular Vote2:14,037
Percentage2:40.07%
Map Size:260px
President
Before Election:Andrew Jackson
Before Party:Democratic Party (United States)
Before Color:FF3333
After Election:Martin Van Buren
After Party:Democratic Party (United States)
After Color:FF3333

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

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 successfully defeated Van Buren.

See also