1852 United States presidential election in Rhode Island

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United States presidential election in Rhode Island, 1852

← 1848 November 2, 1852 1856 →
  Franklin Pierce by Brady.jpg Winfield Scott by Fredricks, 1862.jpg
Nominee Franklin Pierce Winfield Scott
Party Democratic Whig
Home state New Hampshire New Jersey
Running mate William R. King William Alexander Graham
Electoral vote 4 0
Popular vote 8,735 7,626
Percentage 51.37% 44.85%

President before election

Millard Fillmore
Whig

Elected President

Franklin Pierce
Democratic

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

Rhode Island voted for the Democratic candidate, Franklin Pierce, over the Whig Party candidate, Winfield Scott. Pierce won the state by a margin of 6.52%.

This would be the final time until 1912 that a Democratic presidential candidate was able to win Rhode Island and the final time until 1928 that a Democratic candidate won a majority of the popular vote.

Results[edit]

United States presidential election in Rhode Island, 1852[1]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Democratic Franklin Pierce of New Hampshire William Rufus DeVane King of Alabama 8,735 51.37% 4 100.00%
Whig Winfield Scott of New Jersey William Alexander Graham of North Carolina 7,626 44.85% 0 0.00%
Free Soil John Parker Hale of New Hampshire George Washington Julian of Indiana 644 3.79% 0 0.00%
Total 17,005 100.00% 4 100.00%

References[edit]

  1. ^ "1852 Presidential General Election Results - Rhode Island".