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.
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 County Results
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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 was the first of three times that the state voted differently than Massachusetts (along with 1972 and 1980).
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| 1852 United States presidential election in Rhode Island[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% | ||||
See also
editReferences
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