1852 United States presidential election in Rhode Island
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Elections in Rhode Island |
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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%.
Results
United States presidential election in Rhode Island, 1852[1] | ||||||||
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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% |