1836 United States presidential election in Pennsylvania

Summary

The 1836 United States presidential election in Pennsylvania took place between November 3 and December 7, 1836, as part of the 1836 United States presidential election. Voters chose 30 representatives, or electors to the Electoral College, who voted for President and Vice President.

1836 United States presidential election in Pennsylvania

← 1832 November 3 – December 7, 1836 1840 →
 
Nominee Martin Van Buren William Henry Harrison
Party Democratic Whig
Home state New York Ohio
Running mate Richard Johnson Francis Granger
Electoral vote 30 0
Popular vote 91,457 87,235
Percentage 51.18% 48.82%

President before election

Andrew Jackson
Democratic

Elected President

Martin Van Buren
Democratic

Pennsylvania voted for the Democratic candidate, Martin Van Buren, over the Whig candidate, William Henry Harrison. Van Buren won Pennsylvania by a narrow margin of 2.36%. The result would ultimately prove decisive in Van Buren's victory; had Harrison won the state, then Van Buren would not have achieved a majority in the Electoral College, meaning that the election would have been decided in the House of Representatives.

Results edit

1836 United States presidential election in Pennsylvania[1]
Party Candidate Votes Percentage Electoral votes
Democratic Martin Van Buren 91,457 51.18% 30
Whig William Henry Harrison 87,235 48.82% 0
Totals 178,692 100.0% 30

See also edit

References edit

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