1852 United States presidential election in Ohio

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

November 2, 1852

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 Washington Julian Electoral vote 23 0 0 Popular vote 168,933 152,523 31,732 Percentage 47.83% 43.18% 8.98%

County Results Pierce   40-50%   50-60%   60-70%   70-80% Scott   40-50%   50-60%   60-70% Hale   40-50%

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.[lower-alpha 1]

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

Results

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

• United States presidential elections in Ohio

Notes

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

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.