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1840 United States presidential election in Pennsylvania

From Wikipedia, the free encyclopedia

1840 United States presidential election in Pennsylvania

← 1836 October 30 – December 2, 1840 1844 →
William Henry Harrison (cropped).jpg
Van Buren.jpg
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%

President before election

Martin Van Buren

Elected President

William Henry Harrison

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 Whig challenger William Henry Harrison over Democratic incumbent Martin Van Buren by just 334 votes, a margin of 0.12%. It is the narrowest margin of victory in a presidential election in Pennsylvania history, with Donald Trump's 2016 win following close behind at 0.72%.

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1840 United States presidential election in Pennsylvania[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

See also


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

This page was last edited on 26 March 2021, at 18:23
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