United States presidential election in Pennsylvania, 1840

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United States presidential election in Pennsylvania, 1840
Pennsylvania
1836 ←
October 30 – December 2, 1840
→ 1844

  WHenryHarrison.png MartinVanBuren.png
Nominee William Henry Harrison Martin Van Buren
Party Whig Democratic
Home state Ohio New York
Running mate John Tyler none
Electoral vote 30 0
Popular vote 144,010 143,676
Percentage 50.00% 49.88%

The 1840 United States presidential election in Pennsylvania took place between October 30 and December 2, 1840, as part of the 1840 United States presidential election. Voters chose 30 representatives, or electors to the Electoral College, who voted for President and Vice President.

Pennsylvania voted for the Whig candidate, William Henry Harrison, over the Democratic candidate, Martin Van Buren. Harrison won Pennsylvania by a margin of 0.12%.

Results[edit]

United States presidential election in Pennsylvania, 1840[1]
Party Candidate Votes Percentage Electoral votes
Whig William Henry Harrison 144,010 50.00% 30
Democratic Martin Van Buren 143,676 49.88% 0
Liberty James G. Birney 340 0.12% 0
Totals 288,026 100.0% 30

References[edit]

  1. ^ "1840 Presidential General Election Results - Pennsylvania". U.S. Election Atlas. Retrieved 4 August 2012.