1836 United States presidential election in Rhode Island
The 1836 United States presidential election in Rhode Island took place between November 3 and December 7, 1836, as part of the 1836 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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Elections in Rhode Island |
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Rhode Island voted for Democratic candidate Martin Van Buren over Whig candidate William Henry Harrison. Van Buren won Rhode Island by a narrow margin of 4.48%.
This was the first time that Rhode Island ever voted for a Democratic presidential candidate, and Van Buren's performance would not be bettered by a Democrat in Rhode Island until Franklin D. Roosevelt in 1932.[1]
Results
1836 United States presidential election in Rhode Island[2] | ||||||||
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Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
Count | % | Count | % | |||||
Democratic | Martin Van Buren of New York | Richard M. Johnson of Kentucky | 2,964 | 52.24% | 4 | 100.00% | ||
Whig | William Henry Harrison of Ohio | Francis Granger of New York | 2,710 | 47.76% | 0 | 0.00% | ||
Total | 5,674 | 100.00% | 4 | 100.00% |
References
- "Presidential General Election Results Comparison – Rhode Island". Dave Leip’s U.S. Election Atlas. Retrieved 25 October 2019.
- "1836 Presidential General Election Results - Rhode Island". Dave Leip’s U.S. Election Atlas. Retrieved 23 December 2013.
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