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County Results Lincoln 50–60% 60–70%
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The1860 United States presidential election in Rhode Island took place on November 2, 1860, as part of the1860 United States presidential election. Voters chose four electors of theElectoral College, who voted forpresident andvice president.
Rhode Island was won byRepublican candidateAbraham Lincoln, who won by a margin of 22.74%.
| 1860 United States presidential election in Rhode Island[1] | |||||
|---|---|---|---|---|---|
| Party | Candidate | Votes | Percentage | Electoral votes | |
| Republican | Abraham Lincoln | 12,244 | 61.37% | 4 | |
| Fusion | Stephen A. Douglas /John C. Breckinridge /John Bell | 7,707 | 38.63% | 0 | |
| Totals | 19,951 | 100.0% | 4 | ||
| County | Abraham Lincoln Republican | Stephen A. Douglas/ John C. Breckinridge/John BellFusion | Total Votes cast | ||
|---|---|---|---|---|---|
| # | % | # | % | ||
| Bristol | 667 | 59.08% | 462 | 40.92% | 1,129 |
| Kent | 1,246 | 65.48% | 657 | 34.52% | 1,903 |
| Newport | 1,610 | 64.68% | 879 | 35.32% | 2,489 |
| Providence | 7,202 | 59.63% | 4,875 | 40.37% | 12,077 |
| Washington | 1,519 | 64.56% | 834 | 35.44% | 2,353 |
| Total | 12,244 | 61.37% | 7,707 | 38.63% | 19,951 |
With 61.37% of the popular vote, Rhode Island would prove to be Lincoln's fifth strongest state in terms of popular vote percentage in the 1860 election afterVermont,Minnesota, neighboringMassachusetts andMaine.[3]
LikeNew Jersey,New York andPennsylvania, Rhode Island was one of the four states that had a fusion ticket for theDemocrats, which was supported just by the Northern Democrats but also supporters ofSouthern Democrats andConstitutional Unionists. However, unlike in the other three states the electors on the Democratic ticket in Rhode Island were pledged solely to Douglas, therefore some sources credit the Democratic popular vote to the Northern Democratic nominee.
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