1852 United States presidential election in Ohio

Summary

The 1852 United States presidential election in Ohio was held on November 2, 1852 as part of the 1852 United States presidential election. State voters chose 23 electors to the Electoral College, who voted for President and Vice President.

1852 United States presidential election in Ohio

← 1848 November 2, 1852 1856 →
 
Nominee Franklin Pierce Winfield Scott John P. Hale
Party Democratic Whig Free Soil
Home state New Hampshire New Jersey New Hampshire
Running mate William R. King William A. Graham George W. Julian
Electoral vote 23 0 0
Popular vote 168,933 152,523 31,732
Percentage 47.83% 43.18% 8.98%

County Results

President before election

Millard Fillmore
Whig

Elected President

Franklin Pierce
Democratic

Ohio was narrowly won by the Democratic Party candidate, Franklin Pierce, who won the state with a plurality of 47.83 percent of the popular vote. The Whig Party candidate, Winfield Scott, garnered 43.18 percent of the popular vote, and Free Soil Party candidate John P. Hale gained 8.98 percent,[1] a figure exceeded by a third-party candidate in Ohio only six times since.[a]

Pierce would become the final Democrat to win Ohio until Woodrow Wilson won it in 1912.

Results edit

1852 United States presidential election in Ohio
Party Candidate Votes Percentage Electoral votes
Democratic Franklin Pierce 168,933 47.83% 23
Whig Winfield Scott 152,523 43.18% 0
Free Soil John P. Hale 31,732 8.98% 0
Totals 353,188 100.0% 23

See also edit

Notes edit

  1. ^ In 1912, both Theodore Roosevelt and Eugene V. Debs exceeded Hale’s share, as did Robert M. La Follette in 1924, George Wallace in 1968, and Ross Perot in both 1992 and 1996.[2]

References edit

  1. ^ "1852 Presidential General Election Results – Ohio". Dave Leip’s U.S. Election Atlas.
  2. ^ "Presidential General Election Results Comparison – Ohio". Dave Leip’s U.S. Election Atlas.