1836 United States presidential election in Rhode Island

1836 United States presidential election in Rhode Island
	November 23, 1836
Van Buren 50–60%	Harrison 50–60%
President before election; Andrew Jackson; Democratic	Elected President ; Martin Van Buren; Democratic

A presidential election was held in Rhode Island on November 23, 1836 as part of the 1836 United States presidential election.^[1] Voters chose four representatives, or electors to the Electoral College, who voted for President and Vice President.

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.^[2]

Results

1836 United States presidential election in Rhode Island^[3]
Party		Candidate	Running mate	Popular vote		Electoral vote
Party		Candidate	Running mate	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 Elections". Weekly Messenger. November 12, 1836.
^ "Presidential General Election Results Comparison – Rhode Island". Dave Leip’s U.S. Election Atlas. Retrieved October 25, 2019.
^ "1836 Presidential General Election Results - Rhode Island". Dave Leip’s U.S. Election Atlas. Retrieved December 23, 2013.

[1] "Presidential Elections". Weekly Messenger. November 12, 1836.

[2] "Presidential General Election Results Comparison – Rhode Island". Dave Leip’s U.S. Election Atlas. Retrieved October 25, 2019.

[results-3] "1836 Presidential General Election Results - Rhode Island". Dave Leip’s U.S. Election Atlas. Retrieved December 23, 2013.

[1]

[2]

[3]

Results

See also

References