1852 United States presidential election in Vermont

The 1852 United States presidential election in Vermont took place on November 2, 1852, as part of the 1852 United States presidential election. Voters chose five representatives, or electors to the Electoral College, who voted for President and Vice President.

1852 United States presidential election in Vermont

← 1848 November 2, 1852 1856 →

Nominee Winfield Scott Franklin Pierce John P. Hale Party Whig Democratic Free Soil Home state New Jersey New Hampshire New Hampshire Running mate William A. Graham William R. King George W. Julian Electoral vote 5 0 0 Popular vote 22,173 13,044 8,621 Percentage 50.52% 29.72% 19.64%

County Results

President before election Millard Fillmore Whig Elected President Franklin Pierce Democratic

Vermont voted for the Whig Party candidate, Winfield Scott, over the Democratic candidate, Franklin Pierce. Scott won the state by a margin of 20.8%. This would be the last election until 1916 where a Democratic candidate won more than 40% of the vote in a Vermont county.

Vermont was one of only four states to vote for Scott. The other three were neighboring Massachusetts, Kentucky, and Tennessee.

With 50.52% of the popular vote, Vermont would prove to be Scott's third strongest state after Kentucky and Tennessee.^[1] It was also Scott's strong state by margin by victory.

Free Soil Party candidate John P. Hale won 19.64% of the popular vote in the state, making Vermont his second strongest state after neighboring Massachusetts. Lamoille County was won by Hale, which would make him the final third-party candidate to win a single county in Vermont until Progressive Party candidate Theodore Roosevelt won six counties in 1912.

After this election, Vermont would vote for the nominee of the Republican Party, which would be formed in 1854, for 27 consecutive elections, from 1856 through 1960—as of 2020, still the most of any state.^[2]