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1832 United States presidential election in Ohio

From Wikipedia, the free encyclopedia

1832 United States presidential election in Ohio

← 1828 November 2 – December 5, 1832 1836 →
 
Andrew Jackson.jpg
Henry Clay.JPG
Nominee Andrew Jackson Henry Clay
Party Democratic National Republican
Home state Tennessee Kentucky
Running mate Martin Van Buren John Sergeant
Electoral vote 21 0
Popular vote 81,246 76,539
Percentage 51.33% 48.35%

The 1832 United States presidential election in Ohio took place between November 2 and December 5, 1832, as part of the 1832 United States presidential election. Voters chose 21 representatives, or electors to the Electoral College, who voted for President and Vice President.

Ohio voted for the Democratic Party candidate, Andrew Jackson, over the National Republican candidate, Henry Clay, and the Anti-Masonic Party candidate, William Wirt. Jackson won Ohio by a margin of 2.98%.

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  • ✪ The American Presidential Election of 1860
  • ✪ The American Presidential Election of 1840
  • ✪ The American Presidential Election of 1888
  • ✪ The American Presidential Election of 1880
  • ✪ The American Presidential Election of 1872

Transcription

Results

1832 United States presidential election in Ohio[1]
Party Candidate Votes Percentage Electoral votes
Democratic Andrew Jackson 81,246 51.33% 21
National Republican Henry Clay 76,539 48.35% 0
Anti-Masonic William Wirt 509 0.32% 0
Totals 158,294 100.0% 21

References

  1. ^ "1832 Presidential General Election Results - Ohio". U.S. Election Atlas. Retrieved 12 April 2013.


This page was last edited on 16 September 2019, at 22:35
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