1836 United States presidential election in Rhode Island

Main article: 1836 United States presidential election

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.

1836 United States presidential election in Rhode Island

← 1832 November 3 – December 7, 1836 1840 →

Nominee Martin Van Buren William Henry Harrison Party Democratic Whig Home state New York Ohio Running mate Richard Johnson Francis Granger Electoral vote 4 0 Popular vote 2,964 2,710 Percentage 52.24% 47.76%

County Results Van Buren   50-60% Harrison   50-60%

President before election Andrew Jackson Democratic Elected President Martin Van Buren Democratic

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]
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%

See also

• United States presidential elections in Rhode Island