1836 United States presidential election in Illinois

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

1836 United States presidential election in Illinois

← 1832 November 3 – December 7, 1836 1840 →
  Martin Van Buren circa 1837 crop.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 Johnson Francis Granger
Electoral vote 5 0
Popular vote 18,369 15,220
Percentage 54.69% 45.31%

Illinois voted for the Democratic candidate, Martin Van Buren, over Whig candidate William Henry Harrison. Van Buren won Illinois by a margin of 9.38%.

ResultsEdit

1836 United States presidential election in Illinois[1]
Party Candidate Votes Percentage Electoral votes
Democratic Martin Van Buren 18,369 54.69% 5
Whig William Henry Harrison 15,220 45.31% 0
Totals 33,589 100.0% 5

See alsoEdit

ReferencesEdit

  1. ^ "1836 Presidential General Election Results - Illinois". U.S. Election Atlas. Retrieved 4 August 2012.